9. Properties of Graphs

Exercises

a. Overview

b. Value & Limit Information

For the function \(f(x)=x^4-2x^3-15x^2\), find the intercepts and intervals where the function is positive and negative. Then plot these facts on a numberline.

The \(x\)-intercepts are the points where the function is \(0\).
The \(y\)-intercept is the value of the function at \(x=0\)
Test the function at a point in each interval to find where it is positive or negative.

The \(x\)-intercepts occur at \(x = -3\), \(x = 0\), \(x = 5\).
The \(y\)-intercept occurs at \(y = 0\).
The function is positive on \((-\infty,-3)\) and \((5,\infty)\).
The function is negative on \((-3,0)\) and \((0,5)\).

We plot this information on a numberline:

To find the \(x\)-intercepts, we need to find where the function is \(0\). We solve \(f(x)= 0 \) by factoring: \[\begin{aligned} x^4-2x^3-15x^2&=0 \\ x^2(x^2-2x-15)&=0 \\ x^2(x-5)(x+3)&=0 \end{aligned}\] So the \(x\)-intercepts are \(x=0\), \(x=5\) and \(x=-3\).

To find the \(y\)-intercept, we plug \(x=0\) into \(f(x)=x^4-2x^3-15x^2\): \[ f(0)=0^4-2(0)^3-15(0)^2=0 \] So the \(y\)-intercept is at \(y=0\).

To find when the function is positive or negative, evaluate \(f(x)\) at test values between the \(x\)-intercepts: \[\begin{aligned} f(-4) &= (-4)^4-2(-4)^3-15(-4)^2 = 144 &\gt 0 \\ f(-1) &= (-1)^4-2(-1)^3-15(-1)^2 = -12 &\lt 0 \\ f(1) &= (1)^4-2(1)^3-15(1)^2 = -16 &\lt 0 \\ f(6) &= (6)^4-2(6)^3-15(6)^2 = 324 &\gt 0 \end{aligned}\] So the function is positive on \((-\infty,-3)\) and \((5,\infty)\) and negative on \((-3,0)\) and \((0,5)\).
We plot this information on a numberline:

mjs
For the \(f(x)=\dfrac{x^2-5x+6}{x^2-4}\), find the intercepts and asymptotes. For each vertical asymptote, determine whether the graph goes to \(\infty\) or \(-\infty\) on each side.

Factor the function.
\(x\)-intercepts: Find where the function is \(0\).
\(y\)-intercepts: Compute \(f(0)\).
Horizontal asymptotes: Take the limit as \(x\) approaches \(\pm\infty\).
Vertical asymptotes: Find the places where the function is undefined. Then, take the limit as \(x\) approaches each asymptote from the left and right.

The \(x\)-intercept is \(x=3\). The \(y\)-intercept is \(y=-\,\dfrac{3}{2}\).
The horizontal asymptotes are \(y=1\) at both \(x\to\infty\) and \(x\to-\infty\).

There is a vertical asymptote at \(x=-2\) which has up/down behavior.

Before doing anything, we factor the function and cancel any common factors. \[ f(x)=\dfrac{x^2-5x+6}{x^2-4} = \dfrac{(x-3)(x-2)}{(x-2)(x+2)} = \dfrac{x-3}{x+2} \] except that the function is not defined at \(x=2\). To find the \(x\)-intercept, we set \(f(x)=0\) and solve: \[ f(x)=\dfrac{x-3}{x+2}=0 \implies x-3=0 \implies x=3 \]

To find the \(y\)-intercept, we set \(x=0\) in \(f(x)\): \[ f(0)=\dfrac{0-3}{0+2} = -\,\dfrac{3}{2} \]

To find the horizantal asymptotes, we take the limits as \(x\to \infty\) and \(x \to -\infty\). First, as \(x \to \infty\): \[\begin{aligned} \lim_{x\to\infty} &f(x) = \lim_{x\to\infty} \dfrac{x-3}{x+2} \\ &=\lim_{x\to\infty}\dfrac{\;\rule[-8pt]{0pt}{10pt}1-\dfrac{3}{x}\;} {\rule{0pt}{15pt}1+\dfrac{2}{x}} = \dfrac{1-0}{1+0} = 1 \end{aligned}\] Similarly, as \(x \to -\infty\): \[\begin{aligned} \lim_{x\to -\infty} &f(x) = \lim_{x\to -\infty} \dfrac{x-3}{x+2}= 1 \end{aligned}\] So the horizantal asymptote is \(y = 1\) at both \(x\to \infty\) and \(x \to -\infty\).

To find the vertical asymptotes, we set the denominator, \(x+2\), equal to \(0\) and solve, which gives \(x=-2\). To find whether the graph is going to \(-\infty\) or \(\infty\) on each side, we take the limit as \(x\) approches \(-2\) from the left and the right. First, from the left: \[\begin{aligned} \lim_{x\to -2^-} &f(x) = \lim_{x\to -2^-} \dfrac{(x-3)}{(x+2)} \\ &= \dfrac{-2^--3}{-2^-+2} = \dfrac{-5^-}{0^-}=\infty \end{aligned}\] This is positive infinity because \(0^-\) and \(-5^-\) are both negative. Now from the right: \[\begin{aligned} \lim_{x\to -2^+} &f(x) = \lim_{x\to -2^+} \dfrac{(x-3)}{(x+2)} \\ &= \dfrac{-2^+-3}{-2^++2} = \dfrac{-5^+}{0^+}=-\infty \end{aligned}\] This is negative infinity because \(0^+\) is positive while \(-5^+\) is still negative. So the aysmptote at \(x=-2\) has up/down behavior.

mjs
Find the slant asymptotes of the function \(g(x)=\dfrac{6x^4-4x^3+7}{2x^3+4x}\).

A slant asymptote satisfies: \(\displaystyle \lim_{x\to \pm\infty} g(x) - (mx+b) = 0\).
The coefficients are found using: \(\displaystyle m_{\pm} = \lim_{x \to \pm \infty} \dfrac{g(x)}{x} \) and \(\displaystyle b_{\pm} = \lim_{x \to \pm \infty} g(x) - m_{\pm}x \)

The slant asymptote is \(y=3x-2\).

To find the slant asymptote as \(x \to \infty\), we compute: \[\begin{aligned} m_+&=\lim_{x\to \infty} \dfrac{g(x)}{x} \\ &=\lim_{x\to \infty} \dfrac{1}{x}\left(\dfrac{6x^4-4x^3+7}{2x^3+4x}\right) \\ &=\lim_{x\to \infty} \left(\dfrac{6x^4-4x^3+7}{2x^4+4x^2}\right)=3 \end{aligned}\] and \[\begin{aligned} b_+&=\lim_{x\to \infty} g(x)-3x \\ &=\lim_{x\to \infty} (\dfrac{6x^4-4x^3+7}{2x^3+4x}-3x) \\ &=\lim_{x\to \infty} (\dfrac{6x^4-4x^3+7-6x^4-12x^2}{2x^3+4x}) \\ &=\lim_{x\to \infty} (\dfrac{-4x^3+7-12x^2}{2x^3+4x})=-2 \end{aligned}\] So the slant asymptote as \(x \to \infty\) is \(y=3x-2\).

To find the slant asymptote as \(x \to -\infty\), the details of the limit computations are identical. So: \[\begin{aligned} m_-&=\lim_{x\to -\infty} \dfrac{g(x)}{x}=3\\ b_-&=\lim_{x\to -\infty} g(x)-3x=-2 \end{aligned}\] So the slant asymptote as \(x \to -\infty\) is \(y=3x-2\).

To check this we look at a plot of \(g(x)\) and \(y=3x-2\).

mjs
Find the slant asymptotes of the function \(h(x)=\dfrac{x^2|4x-5|}{x^2+5}\).

For large positive \(x\), what is \(|4x-5|\)?
For large negative \(x\), what is \(|4x-5|\)?

A slant asymptote satisfies: \(\displaystyle \lim_{x\to \pm\infty} h(x) - (mx+b) = 0\).
The coefficients are found using: \(\displaystyle m_{\pm} = \lim_{x \to \pm \infty} \dfrac{h(x)}{x} \) and \(\displaystyle b_{\pm} = \lim_{x \to \pm \infty} h(x) - m_{\pm}x \)

As \(x \to \infty\) the slant asymptote is \(y=4x-5\) and
as \(x \to -\infty\) the slant asymptote is \(y=-4x+5\).

For large positive \(x\), we have \(|4x-5|=4x-5\). So to find the slant asymptote to \(h(x)\) as \(x\to\infty\), we compute: \[\begin{aligned} m_+&=\lim_{x\to \infty} \dfrac{h(x)}{x} =\lim_{x\to \infty} \dfrac{1}{x}\left(\dfrac{x^2|4x-5|}{x^2+5}\right) \\ &=\lim_{x\to \infty} \dfrac{1}{x}\left(\dfrac{x^2(4x-5)}{x^2+5}\right) \\ &=\lim_{x\to \infty} \left(\dfrac{4x^3-5x^2}{x^3+5x}\right)=4 \\ \end{aligned}\] and \[\begin{aligned} b_+&=\lim_{x\to \infty} (h(x)-4x) =\lim_{x\to \infty} \left(\dfrac{x^2(4x-5)}{x^2+5}-4x\right)\\ &=\lim_{x\to \infty} \left(\dfrac{(4x^3-5x^2)-(4x^3+20x)}{x^2+5}\right)\\ &=\lim_{x\to \infty} \dfrac{-5x^2-20x}{x^2+5} = -5 \end{aligned}\] So the slant asymptote as \(x \to \infty\) is \(y=4x-5\).

On the other hand, for large negative \(x\), we have \(|4x-5|=-4x+5\). So as \(h(x)\) goes to \(-\infty\), we compute: \[\begin{aligned} m_-&=\lim_{x\to -\infty} \dfrac{h(x)}{x} =\lim_{x\to -\infty} \dfrac{1}{x}\left(\dfrac{x^2(-4x+5)}{x^2+5}\right) \\ &=\lim_{x\to -\infty} \dfrac{-4x^3+5x^2}{x^3+5x} = -4 \end{aligned}\] and \[\begin{aligned} b_-&=\lim_{x\to -\infty} (h(x)+4x) =\lim_{x\to -\infty} (\dfrac{x^2|4x-5|}{x^2+5}+4x) \\ &=\lim_{x\to -\infty} (\dfrac{-4x^3+5x^2+4x^3+20x}{x^2+5})\\ &=\lim_{x\to -\infty} \dfrac{5x^2+20x}{x^2+5} = 5 \end{aligned}\] So the slant asymptote as \(x \to -\infty\) is \(y=-4x+5\).

mjs

To check this we look at a plot of \(h(x)=\dfrac{x^2|4x-5|}{x^2+5}\),
\(y=4x-5\) - the asymptote as \(x\to\infty\) and
\(y=-4x+5\) - the asymptote as \(x\to-\infty\).
Find the slant asymptotes of the function \(k(x)=\dfrac{(4x+2)e^x+9x}{2e^x-3}\).

Recall \(\displaystyle \lim_{x\to\infty} e^x=\infty\) but \(\displaystyle \lim_{x\to-\infty} e^x=0\).

A slant asymptote satisfies: \(\displaystyle \lim_{x\to \pm\infty} k(x) - (mx+b) = 0\).
The coefficients are found using: \(\displaystyle m_{\pm} = \lim_{x \to \pm \infty} \dfrac{k(x)}{x} \) and \(\displaystyle b_{\pm} = \lim_{x \to \pm \infty} k(x) - m_{\pm}x \)

The slant aysmptote as \(x \to \infty\) is \(y=2x+1\) and
as \(x \to -\infty\) the slant asymptote is \(y=-3x\).

To find the slant asymptote to \(k(x)\) as \(x\to\infty\), we compute: \[\begin{aligned} m_+&=\lim_{x \to \infty} \dfrac{k(x)}{x} =\lim_{x \to \infty} \dfrac{1}{x}\left(\dfrac{(4x+2)e^x+9x}{2e^x-3}\right) \\ &=\lim_{x \to \infty} \dfrac{(4x+2)e^x+9x}{2xe^x-3x} \\ &=\lim_{x \to \infty} \dfrac{4+\dfrac{2}{x}+9e^{-x}}{2-3e^{-x}} =\dfrac{4}{2}=2 \end{aligned}\] and \[\begin{aligned} b_+&=\lim_{x \to\infty}(k(x) - 2x) \\ &=\lim_{x \to\infty} \dfrac{(4x+2)e^x+9x}{2e^x-3} - 2x \\ &=\lim_{x \to\infty} \dfrac{4xe^x+2e^x+9x-4xe^x+6x}{2e^x-3} \\ &=\lim_{x \to\infty} \dfrac{2e^x+15x}{2e^x-3} = 1 \end{aligned}\] So the slant asymptote as \(x \to \infty\) is \(y=2x+1\).

To find the slant asymptote to \(k(x)\) as \(x\to-\infty\), we compute: \[\begin{aligned} m_-&=\lim_{x \to - \infty} \dfrac{k(x)}{x} =\lim_{x \to - \infty} \dfrac{1}{x}\left(\dfrac{(4x+2)e^x+9x}{2e^x-3}\right) \\ &=\lim_{x \to - \infty} \dfrac{4xe^x+2e^x+9x}{2xe^x-3x} \\ &=\lim_{x \to - \infty} \dfrac{4e^x+\dfrac{2e^x}{x}+9}{2e^x-3} =\dfrac{9}{-3}=-3 \end{aligned}\] and \[\begin{aligned} b_-&=\lim_{x \to - \infty} (k(x) + 3x) \\ &=\lim_{x \to - \infty} \dfrac{(4x+2)e^x+9x}{2e^x-3} + 3x \\ &=\lim_{x \to -\infty} \dfrac{4xe^x+2e^x+9x+6xe^x-9x}{2e^x-3} \\ &=\lim_{x \to -\infty} \dfrac{10xe^x+2e^x}{2e^x-3} =\dfrac{0}{-3} = 0 \end{aligned}\] So the slant asymptote as \(x \to -\infty\) is \(y=-3x\).

mjs

To check this we look at a plot of \(k(x)=\dfrac{(4x+2)e^x+9x}{2e^x-3}\),
\(y=2x+1\) - the asymptote as \(x\to\infty\) and
\(y=-3x\) - the asymptote as \(x\to-\infty\).

For the function \(f(x)=\dfrac{x^4+5x^2-2}{x^3}\), find the intercepts and asymptotes. For each vertical asymptote, determine whether the graph goes to \(\infty\) or \(-\infty\) on each side.

The \(x\)-intercepts are \(x=\pm\sqrt{\dfrac{-5+\sqrt{33}}{2}}\approx\pm0.61\). There are no \(y\)-intercepts as there is a vertical asymptote at \(x=0\) with up/down behavior. There is a slant asymptote, \(y=x\), as \(x\to\infty\) and \(x\to-\infty\).

First, we find the zeros of the function by setting \(f(x)\) equal to \(0\): \[\begin{aligned} x^4&+5x^2-2=0 \\ x^2&=\dfrac{-5\pm\sqrt{5^2-4(1)(-2)}}{2(1)} \\ &=\dfrac{-5\pm\sqrt{33}}{2} \approx .372, -5.37 \end{aligned}\] Only the positive solution can be a square. So the \(x\)-intercept is: \[ x=\pm\sqrt{\dfrac{-5+\sqrt{33}}{2}} \approx \pm\sqrt{.372} \approx \pm.61 \]

We notice that if \(x=0\) the denominator of our function is \(0\) which means there is a vertical asymptote at \(x=0\) and no \(y\)-intercepts.

To check what the function does on either side of the asymptote we take the limit as \(x\to0\) from the left and right. First from the left: \[\begin{aligned} \lim_{x \to 0^-} f(x) &=\lim_{x \to 0^-} \dfrac{x^4+5x^2-2}{x^3} \\ &=\dfrac{(0^-)^4+5(0^-)^2-2}{(0^-)^3} \\ &=\dfrac{-2}{0^-} =\infty \end{aligned}\] Then from the right: \[\begin{aligned} \lim_{x \to 0^+} f(x) &=\lim_{x \to 0^+} \dfrac{x^4+5x^2-2}{x^3} \\ &=\dfrac{(0^+)^4+5(0^+)^2-2}{(0^+)^3} \\ &=\dfrac{-2}{0^+} =-\infty \end{aligned}\] So the vertical asymptote at \(x=0\) has up/down behavior.

Since the function is infinite at both ends, we check for slant asymptotes. First as \(x\to\infty\): \[\begin{aligned} m_+&=\lim_{x \to \infty} \dfrac{f(x)}{x} \\ &=\lim_{x \to \infty} \dfrac{1}{x}\left(\dfrac{x^4+5x^2-2}{x^3}\right) \\ &=\lim_{x \to \infty} \left(1+\dfrac{5}{x^2}-\dfrac{2}{x^4}\right) = 1 \end{aligned}\] and \[\begin{aligned} b_+&=\lim_{x \to\infty}(f(x) - x) \\ &=\lim_{x \to\infty} \dfrac{x^4+5x^2-2}{x^3} - x \\ &=\lim_{x \to\infty} \dfrac{x^4+5x^2-2-x^4}{x^3} \\ &=\lim_{x \to \infty} \dfrac{5}{x}-\dfrac{2}{x^3} = 0 \end{aligned}\] So the slant asymptote as \(x \to \infty\) is \(y=x\).

Similarly, we find the slant asymptote as \(x\to-\infty\) \[\begin{aligned} m_-&=\lim_{x \to -\infty} \dfrac{f(x)}{x} \\ &=\lim_{x \to -\infty} \dfrac{1}{x}\left(\dfrac{x^4+5x^2-2}{x^3}\right) \\ &=\lim_{x \to -\infty} 1+\dfrac{5}{x^2}-\dfrac{2}{x^4} = 1 \end{aligned}\] and \[\begin{aligned} b_-&=\lim_{x \to\infty}(f(x) - x) \\ &=\lim_{x \to-\infty} \dfrac{x^4+5x^2-2}{x^3} - x \\ &=\lim_{x \to-\infty} \dfrac{x^4+5x^2-2-x^4}{x^3} \\ &=\lim_{x \to -\infty} \dfrac{5}{x}-\dfrac{2}{x^3} = 0 \end{aligned}\] So the slant asymptote as \(x\to\infty\) and \(x\to-\infty\) is \(y=x\).

All of this agrees with the actual plot.

mjs

c. First Derivative Information

For the function \(f(x)=\dfrac{1}{6}x^6-\dfrac{4}{5}x^5+x^4\), find the critical points, the intervals where the function is increasing, and decreasing and the local minima, local maxima and horizontal inflection points. Then plot these facts on a numberline.

The critical points are \(x=0\) and \(x=2\). The function is decreasing from \(-\infty\) to \(0\), and incresing from \(0\) to \(\infty\). There is a local minima at \(x=0\) and a horizantal inflection point at \(x=2\).

We plot this information on a numberline:

To find the critical points, we take the derivative of \(f(x)=\dfrac{1}{6}x^6-\dfrac{4}{5}x^5+x^4\), set it equal to 0 and solve for \(x\): \[\begin{aligned} f'(x)=x^5-4x^4+4x^3 &= 0 \\ x^3(x^2-4x+4) &= 0 \\ x^3(x-2)^2 &= 0 \\ x &= 0, 2 \end{aligned}\] To find where the function is increasing or decreasing, we test values on each side of the critcal points: \[\begin{aligned} f'(-1)&=(-1)^5-4(-1)^4+4(-1)^3 \!\!\!\!\!\!&&= -9 \lt 0 \\ f'(1)&=(1)^5-4(1)^4+4(1)^3 \!\!\!\!\!\!&&= 1 \gt 0 \\ f'(3)&=(3)^5-4(3)^4+4(3)^3 \!\!\!\!\!\!&&= 27 \gt 0 \\ \end{aligned}\] Therefore, the function is decreasing from \(-\infty\) to \(0\), and increasing from \(0\) to \(\infty\). Since the function goes from decreasing to increasing at \(x=0\), it is a local minima. At \(x=2\), the function stays increasing so it is a horizantal inflection point.

We plot this information on a numberline:

mjs

For the function \(f(x)=\dfrac{x^4+5x^2-2}{x^3}\) find the critical points, the intervals where the function is increasing, and decreasing and the local minima, local maxima and horizontal inflection points. Then plot these facts on a numberline.

The critical points are \(x=\pm\sqrt{2}\) and \(x=\pm\sqrt{3}\).
The function is increasing on the intervals \((-\infty,-\sqrt{3})\), \((-\sqrt{2},0)\), \((0,\sqrt{2})\) and \((\sqrt{3},\infty)\),
and decreasing on \((-\sqrt{3},-\sqrt{2})\), and \((\sqrt{2},\sqrt{3})\).
There are local minima at \(x=-\sqrt2,\sqrt3\) and local maxima at \(x=-\sqrt3,\sqrt2\).
We summarize these results on a number line:

To compute the derivative of \(f(x)\), we could use the Quotient Rule. However, it is easier to first simplify the function: \[\begin{aligned} f(x)&=x+\dfrac{5}{x}-\dfrac{2}{x^3} \\ f'(x)&=1-\dfrac{5}{x^2}+\dfrac{6}{x^4} \\ &=\dfrac{x^4-5x^2+6}{x^4} \\ \end{aligned}\] We set the derivative equal to \(0\) to find our critical points: \[\begin{aligned} f'(x)=&\dfrac{(x^2-2)(x^2-3)}{x^4}=0 \\ x^2 = 2 &\quad \text{or} \quad x^2=3 \\ x = \pm\sqrt{2} &\quad \text{or} \quad x=\pm\sqrt{3} \end{aligned}\] The function values are: \[\begin{aligned} f(-\sqrt{3})&= -\dfrac{22\sqrt{3}}{9} \approx -4.233\\ f(-\sqrt{2})&= -3\sqrt{2} \approx -4.242\\ f(\sqrt{2})&= 3\sqrt{2} \approx 4.242\\ f(\sqrt{3})&= \dfrac{22\sqrt{3}}{9} \approx 4.233 \end{aligned}\] So our critical points are \(\left(\pm\sqrt{2},\pm4.242\right)\) and \(\left(\pm\sqrt{3},\pm4.233\right)\). In addition there is a vertical asymptote at \(x=0\). So we need to check for increasing and decreasing on \(6\) intervals: \[\begin{aligned} (-\infty,-\sqrt{3}), (-\sqrt{3},&-\sqrt{2}), (-\sqrt{2},0), \\ (0,\sqrt{2}), (\sqrt{2},&\sqrt{3}), (\sqrt{3},\infty) \end{aligned}\] We test a point in each interval to see if \(f'(x)\) is positive or negative. \[\begin{aligned} f'(-2)&=\dfrac{(4-2)(4-3)}{16}=\dfrac{1}{8} \gt 0 \\ f'(-1.5)&=\dfrac{(2.25-2)(2.25-3)}{5. 0625} \lt 0 \\ f'(-1)&=\dfrac{(1-2)(1-3)}{1}=2 \gt 0 \\ f'(1)&=\dfrac{(1-2)(1-3)}{1}=2 \gt 0 \\ f'(1.5)&=\dfrac{(2.25-2)(2.25-3)}{5. 0625} \lt 0 \\ f'(2)&=\dfrac{(4-2)(4-3)}{16}=\dfrac{1}{8} \gt 0 \end{aligned}\] So \(f(x)\) is increasing on the intervals \((-\infty,-\sqrt{3})\), \((-\sqrt{2},0)\), \((0,\sqrt{2})\) and \((\sqrt{3},\infty)\), while it is decreasing on \((-\sqrt{3},-\sqrt{2})\), and \((\sqrt{2},\sqrt{3})\). Using the First Derivative Test, we see there are local minima at \(x=-\sqrt2,\sqrt3\) and local maxima at \(x=-\sqrt3,\sqrt2\).
We summarize these results on a number line:

All of this agrees with the actual plot.

mjs

d. Second Derivative Information

For the function \(f(x)=\dfrac{1}{6}x^6-\dfrac{4}{5}x^5+x^4\), find the secondary critical points, the intervals where the function is concave up and concave down, local minima, local maxima and inflection points. Then plot these facts on a numberline.

The secondary critical points are \(x=0\), \(x=\dfrac{6}{5}\), and \(x=2\). The function is concave up on \((-\infty,0)\), \(\left(0,\dfrac{6}{5}\right)\) and \((2,\infty)\) and concave down on \(\left(\dfrac{6}{5},2\right)\). A local minima occurs at \(x=0\), there is no local maxima. \(x=\dfrac{6}{5}\) and \(x=2\) are inflection points.

To find the critical points and secondary critical points, we factor the first and second derivatives and set them equal to \(0\): \[\begin{aligned} f'(x)&=x^5-4x^4+4x^3 \\ &=x^3(x-2)^2=0 \end{aligned}\] The critical points are \(x=0\) and \(x=2\). \[\begin{aligned} f''(x)&=5x^4-16x^3+12x^2 \\ &=x^2(5x^2-16x+12)= 0\\ \end{aligned}\] The solutions are \(x=0\) and \[\begin{aligned} x&=\dfrac{16\pm\sqrt{16^2-4(5)(12)}}{2(5)}\\ &=\dfrac{16\pm\sqrt{16}}{10} =\dfrac{16\pm4}{10} \\ x&=2 \quad \text{and} \quad x=\dfrac{6}{5} \end{aligned}\] So, the secondary critical points are \(x=0,\dfrac{6}{5},2\). To find concavity we plug in test values between these critical points: \[\begin{aligned} f''(-1)&=5(-1)^4-16(-1)^3+12(-1)^2 &&\!\!\!\!\!\!= 33 \gt 0\\ f''(1)&=5(1)^4-16(1)^3+12(1)^2 &&\!\!\!\!\!\!= 1\gt0\\ f''\left(\dfrac{3}{2}\right) &=5\left(\dfrac{3}{2}\right)^4-16\left(\dfrac{3}{2}\right)^3 +12\left(\dfrac{3}{2}\right)^2 &&\!\!\!\!\!\!=-\,\dfrac{27}{16}\lt0\\ f''(3)&=5(3)^4-16(3)^3+12(3)^2 &&\!\!\!\!\!\!= 81\gt0 \end{aligned}\] So, the function is concave up on \((-\infty,0)\), \(\left(0,\dfrac{6}{5}\right)\) and \((2,\infty)\) and concave down on \(\left(\dfrac{6}{5},2\right)\). Since the concavity changes at \(x=\dfrac{6}{5}\) and \(x=2\), they are both inflection points.

We examime the critical points for maxima and minima. We already know \(x=2\) is an inflection point. We can't use the second derivative test on \(x=0\) as \(f''(0) = 0\) so the test fails. Therefore, we need to go to the first derivative to see what is happening at \(x=0\): \[\begin{aligned} f'(-1)&=(-1)^5-4(-1)^4+4(-1)^3 &&\!\!\!\!\!\!\!=-9\lt0 \\ f'(1)&=(1)^5-4(1)^4+4(1)^3 &&\!\!\!\!\!\!\!= 1\gt0 \end{aligned}\] This tells us that the graph goes from decreasing to increasing at \(x=0\) so it's a local minima. We plot this information on a numberline:

All of this agrees with the actual plot.

mjs

For the function \(f(x)=\dfrac{x^4+5x^2-2}{x^3}\), find the secondary critical points, the intervals where the function is concave up and concave down, local minima, local maxima and inflection points. Then plot these facts on a numberline.

The function has a vertical asymptote at \(x=0\).
The secondary critical points are \(x=\pm2\sqrt{\dfrac{3}{5}}\).
The function is concave down on \(\left(-\infty, -2\sqrt{\dfrac{3}{5}}\right)\) and \(\left(0,2\sqrt{\dfrac{3}{5}}\right)\) and concave up on \(\left(-2\sqrt{\dfrac{3}{5}},0\right)\) and \(\left(2\sqrt{\dfrac{3}{5}}, \infty \right)\).
The maxima occur at \(x=-\sqrt{3}\) and \(x=\sqrt{2}\) and the minima occur at \(x=-\sqrt{2}\) and \(x=\sqrt{3}\).
The inflection points are \(x=\pm2\sqrt{\dfrac{3}{5}}\).

We plot this information on a numberline:

From a previous problem we know the critical points are at \(x=\pm\sqrt{3}\) and \(x=\pm\sqrt{2}\).

Similarly we take the second derivative to find the secondary critical points: \[\begin{aligned} f'(x)&=1-\dfrac{5}{x^2}+\dfrac{6}{x^4} \\ f''(x)&=\dfrac{10}{x^3}-\dfrac{24}{x^5} =\dfrac{10x^2-24}{x^5}=0 \\ x^2&=\dfrac{24}{10}=\dfrac{12}{5} \\ x&=\pm\sqrt{\dfrac{12}{5}} \end{aligned}\] So our secondary critical points are \(x=\pm\sqrt{\dfrac{12}{5}}\approx\pm1.55\).

Since there is a vertical asymptote at \(x=0\), we need to check for concavity on the intervals: \[ \left(-\infty, -\sqrt{\dfrac{12}{5}}\right), \left(-\sqrt{\dfrac{12}{5}},0\right), \left(0,\sqrt{\dfrac{12}{5}}\right), \left(\sqrt{\dfrac{12}{5}}, \infty \right) \] We test values of \(f''(x)=\dfrac{10}{x^3}-\dfrac{24}{x^5}\) on each interval: \[\begin{aligned} f''(-2)&=\dfrac{10}{(-2)^3}-\dfrac{24}{(-2)^5} &&= \dfrac{-1}{2} \lt 0 \\ f''(-1)&=\dfrac{10}{(-1)^3}-\dfrac{24}{(-1)^5} &&= 14 \gt 0 \\ f''(1)&=\dfrac{10}{(1)^3}-\dfrac{24}{(1)^5} &&= -14 \lt 0 \\ f''(2)&=\dfrac{10}{(2)^3}-\dfrac{24}{(2)^5} &&= \dfrac{1}{2} \gt 0 \end{aligned}\] So the function is concave down on \(\left(-\infty, -\sqrt{\dfrac{12}{5}}\right)\) and \(\left(0,\sqrt{\dfrac{12}{5}}\right)\) and concave up on \(\left(-\sqrt{\dfrac{12}{5}},0\right)\) and \(\left(\sqrt{\dfrac{12}{5}}, \infty \right)\). Since the concavity changes at \(x=\pm\sqrt{\dfrac{12}{5}}\), these are the inflection points.

We apply the Second Derivative Test to find the local maxima and minima. So we evaluate the second derivative at each critical point: \[\begin{aligned} f''(-\sqrt{3})&=\dfrac{10}{(-\sqrt{3})^3}-\dfrac{24}{(-\sqrt{3})^5} &&= \dfrac{-2\sqrt{3}}{9} \lt 0 \\ f''(-\sqrt{2})&=\dfrac{10}{(-\sqrt{2})^3}-\dfrac{24}{(-\sqrt{2})^5} &&= \dfrac{\sqrt{2}}{2} \gt 0 \\ f''(\sqrt{2})&=\dfrac{10}{(\sqrt{2})^3}-\dfrac{24}{(\sqrt{2})^5} &&= -\dfrac{\sqrt{2}}{2} \lt 0 \\ f''(\sqrt{3})&=\dfrac{10}{(\sqrt(3))^3}-\dfrac{24}{(\sqrt(3))^5} &&= \dfrac{2\sqrt{3}}{9} \gt 0 \end{aligned}\] Since the concavity is positive at \(x=-\sqrt{2}\) and \(x=\sqrt{3}\), these are the minima. Since the concavity is negative at \(x=-\sqrt{3}\) and \(x=\sqrt{2}\), these are the maxima. This agrees with our results from the First Derivative Test.

We plot this information on a numberline. The critical points are open circles. The secondary critical points are closed circles.

All of this agrees with the actual plot.

mjs
Find where each function is concave up or concave down.
1. \(g(x)=\dfrac{x^2}{1+x^2}\)
  
  Find the secondary critical points and find the concavity of the function between them.
  
  \(g(x)\) is concave down on \(\left(-\infty, -\,\dfrac{1}{\sqrt{3}}\right)\) and \(\left(\dfrac{1}{\sqrt{3}},\infty\right)\) and concave up on \(\left(-\,\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)\).
  
  To find where the function is concave up or down, we look at the secondary critical points. First we find the second derivative. \[\begin{aligned} g(x)&=\dfrac{x^2}{1+x^2} \\ g'(x)&=\dfrac{(1+x^2)2x-x^2(2x)}{(1+x^2)^2} =\dfrac{2x}{(1+x^2)^2} \\ g''(x)&=\dfrac{(1+x^2)^2 2-2x2(1+x^2)(2x)}{(1+x^2)^4} \\ &=\dfrac{(1+x^2) 2-2x(2(2x))}{(1+x^2)^3} \\ &=\dfrac{2-6x^2}{(1+x^2)^3} \end{aligned}\] Next we set the second derivative equal to \(0\) and solve for the secondary critical points. \[\begin{aligned} 2-6x^2&=0 \\ x^2&=\dfrac{1}{3} \\ x&= \pm\dfrac{1}{\sqrt{3}} \end{aligned}\] So our secondary critical points are \(x=\pm\dfrac{1}{\sqrt{3}}\). We then plug values on all sides of the critical points to find the concavity in those intervals. \[\begin{aligned} g''(-1)&=\dfrac{2-6}{(1+1)^3} &&=-\dfrac{1}{2} \lt 0\\ g''(0)&=\dfrac{2}{(1)^3} &&=2 \gt 0 \\ g''(1)&=\dfrac{2-6}{(1+1)^3} &&=-\dfrac{1}{2} \lt 0 \\ \end{aligned}\]
  
  Therefore, \(g(x)\) is concave down on \(\left(-\infty, -\,\dfrac{1}{\sqrt{3}} \right)\) and \(\left(\dfrac{1}{\sqrt{3}},\infty\right)\) and concave up on \(\left(-\,\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}}\right)\). We can check this against the graph:
  
  mjs
2. \(g(x)=\dfrac{x}{1+x^2}\)
  
  Find the secondary critical points and find the concavity of the function between them.
  
  \(g(x)\) is concave down on \(\left(-\infty, -\sqrt{3}\right)\) and \(\left(0,\sqrt{3}\right)\) and concave up on \(\left(-\sqrt{3},0\right)\) and \(\left(\sqrt{3},\infty\right)\).
  
  To find where the function is concave up or down, we look at the secondary critical points. First we find the second derivative. \[\begin{aligned} g(x)&=\dfrac{x}{1+x^2} \\ g'(x)&=\dfrac{(1+x^2)(1)-(x)(2x)}{(1+x^2)^2} =\dfrac{1-x^2}{(1+x^2)^2} \\ g''(x)&=\dfrac{(1+x^2)^2(-2x)-(1-x^2)2(1+x^2)(2x)}{(1+x^2)^4} \\ &=\dfrac{(1+x^2)(-2x)-(1-x^2)4x}{(1+x^2)^3} \\ &=\dfrac{2x^3-6x}{(1+x^2)^3} \\ \end{aligned}\] Next we set the second derivative equal to \(0\) and solve for the secondary critical points. \[\begin{aligned} 2x^3-6x &=0 \\ x(x^2-3)&=0 \\ x=0 \quad \text{or} \quad &x=\pm\sqrt{3} \end{aligned}\] So our secondary critical points are \(x=-\sqrt{3}\), \(x=0\) and \(x=\sqrt{3}\). We then plug values on all sides of the critical points to find the concavity in those intervals. \[\begin{aligned} g''(-2)&=\dfrac{2(-2)^3-6(-2)}{(1+(-2)^2)^3} &&= -\,\dfrac{4}{125} \lt 0\\ g''(-1)&=\dfrac{2(-1)^3-6(-1)}{(1+(-1)^2)^3} &&= \dfrac{1}{2} \gt 0\\ g''(1)&=\dfrac{2(1)^3-6(1)}{(1+(1)^2)^3} &&= -\,\dfrac{1}{2} \lt 0\\ g''(2)&=\dfrac{2(2)^3-6(2)}{(1+(2)^2)^3} &&= \dfrac{4}{125} \gt 0\\ \end{aligned}\]
  
  Therefore, \(g(x)\) is concave down on \(\left(-\infty, -\sqrt{3}\right)\) and \(\left(0,\sqrt{3}\right)\) and concave up on \(\left(-\sqrt{3},0\right)\) and \(\left(\sqrt{3},\infty\right)\). We can check this against the graph:
  
  mjs

e. Curve Sketching

Sketch the graph of the function \(f(x)=\dfrac{1}{6}x^6-\dfrac{4}{5}x^5+x^4\).

Use information from the function and the first and the second derivatives to get the local behavior of the graph. Then connect the dots.

to draw a curve between the points following the monotonicity and the concavity and finally to remove the small markings.

To sketch the graph of \(f(x)=\dfrac{1}{6}x^6-\dfrac{4}{5}x^5+x^4\) we begin by finding its \(x\) and \(y\) intercepts. The \(x\) intercepts are the zeros: \[\begin{aligned} f(x)&=\dfrac{1}{6}x^6-\dfrac{4}{5}x^5+x^4 \\ &=\dfrac{x^4}{30}(5x^2-24x+30)=0 \end{aligned}\] There is one \(x\) intercept at \(x=0\). If there were others, they would be \[ x=\dfrac{24\pm\sqrt{24^2-4\cdot5\cdot30}}{10} \] which are imaginary. The \(y\) intercept is the value at \(0\): \[\begin{aligned} f(0)=\dfrac{1}{6}(0)^6-\dfrac{4}{5}(0)^5+(0)^4=0 \end{aligned}\] So, the \(x\) and \(y\) intercepts both occur at the origin, \((x,y)=(0,0)\).

From the first derivative information we know that:
The critical points are \(x=0\) and \(x=2\)
The function decreases on \((-\infty,0)\)
and increases on \((0,2)\) and \((2,\infty)\)
There is a local minimum at \(x=0\)
We get the function values for our critical points \(x=0\) and \(x=2\): \[\begin{aligned} f(0)&= \dfrac{1}{6}(0)^6-\dfrac{4}{5}(0)^5+(0)^4 &&\!\!\!\!\!\!=0 \\ f(2)&= \dfrac{1}{6}(2)^6-\dfrac{4}{5}(2)^5+(2)^4 &&\!\!\!\!\!\!=\dfrac{16}{15} \approx 1.067 \end{aligned}\]

We put all the first derivative information together on a number line

From the second derivative information we know that:
The secondary critical points are \(x=0\), \(x=\dfrac{6}{5}\) and \((2,\dfrac{16}{15})\)
is concave up on \(\left(-\infty,\dfrac{6}{5}\right)\), and \((2,\infty)\)
is concave down on \(\left(\dfrac{6}{5},2\right)\)
The inflection points are \(x=\dfrac{6}{5}\) and \(x=2\).
We find the function values of our secondary critical point: \[ f\left(\dfrac{6}{5}\right)= \dfrac{1}{6}\left(\dfrac{6}{5}\right)^6- \dfrac{4}{5}\left(\dfrac{6}{5}\right)^5+\left(\dfrac{6}{5}\right)^4 =\dfrac{9072}{15625} \approx .581 \]

We put all the second derivative information together on a number line

Then we combine the first and second derivative information and align them vertically so we can draw the graph:

to draw a curve between the points following the monotonicity and the concavity and finally to remove the small markings.

mjs

We check the graph agrees with the properties we found:
\(x\) and \(y\) intercepts at \((0,0)\).
Local minimum at \((0,0)\).
Inflection points at \(\left(\dfrac{6}{5},\dfrac{9072}{15625}\right)\), and \(\left(2,\dfrac{16}{15}\right)\).
It is increasing on \((0,2)\), and \((2,\infty)\).
It is decreasing on \((-\infty,0)\).
It is concave up on \(\left(-\infty,\dfrac{6}{5}\right)\), and \((2,\infty)\).
It is concave down on \(\left(\dfrac{6}{5},2\right)\).
Sketch the graph of the function \(f(x)=\dfrac{x^4+5x^2-2}{x^3}\).

Use information from the function, the first, and the second derivative to get reference points to sketch the graph.

Here is our sketch and graph.

In previous problems, we have found all the information needed to sketch \(\dfrac{x^4+5x^2-2}{x^3}\).

From the function value and limit information we found:
The \(x\)-intercepts are \(x=\dfrac{-5\pm\sqrt{33}}{2} \approx \pm0.61\).
There is no \(y\)-intercept because \(y=0\) is a vertical asymptote.
There is a slant asymptote, \(y=x\), as \(x\to\infty\) and as \(x\to-\infty\).

From the first derivative information we found:
The critical points are \(x=\pm\sqrt{2}\) and \(x=\pm\sqrt{3}\).
The function is increasing on the intervals \((-\infty,-\sqrt{3})\), \((-\sqrt{2},0)\), \((0,\sqrt{2})\) and \((\sqrt{3},\infty)\), while it is decreasing on \((-\sqrt{3},-\sqrt{2})\), and \((\sqrt{2},\sqrt{3})\).
The maxima occur at \(x=-\sqrt{3}\) and \(x=\sqrt{2}\) and the minima occur at \(x=-\sqrt{2}\) and \(x=\sqrt{3}\) with values: \[\begin{aligned} f(-\sqrt{3})&= -\dfrac{22\sqrt{3}}{9} \approx -4.233\\ f(-\sqrt{2})&= -3\sqrt{2} \approx -4.242\\ f(\sqrt{2})&= 3\sqrt{2} \approx 4.242\\ f(\sqrt{3})&= \dfrac{22\sqrt{3}}{9} \approx 4.233 \end{aligned}\]

From the second derivative information we found:
The secondary critical points are \(x=\pm\sqrt{\dfrac{12}{5}}\).
The function is concave down on \(\left(-\infty, -\sqrt{\dfrac{12}{5}}\right)\) and \(\left(0,\sqrt{\dfrac{12}{5}}\right)\) and concave up on \(\left(-\sqrt{\dfrac{12}{5}},0\right)\) and \(\left(\sqrt{\dfrac{12}{5}}, \infty \right)\).
The inflection points are \(x=\pm\sqrt{\dfrac{12}{5}}\) with values: \[\begin{aligned} f\left(-\sqrt{\dfrac{12}{5}}\right)&= -\dfrac{197\sqrt{15}}{180} \approx -4.238\\ f\left(\sqrt{\dfrac{12}{5}}\right)&= \dfrac{197\sqrt{15}}{180} \approx 4.238 \end{aligned}\]
The second derivative test confirmed the maxima and minima.

We put this information together on a number line. The symbols mean:
A solid circle means an \(x\)-intercept.
An empty diamond means a critcal point.
A solid diamond means a secondary critical point.

To sketch the graph, we first plot all the points, the \(x\)-intercepts, the critical points and the secondary critical point, along with what we know about slope and concavity at these points. Finally, to get the final graph, we connect the dots.

mjs
Sketch the graph of the function \(e^{-x^2}\).

Use information from the function, the first, and the second derivative to get reference points to sketch the graph.

Here is our sketch and graph.

To sketch the graph of \(e^{-x^2}\) we first find its intercepts and asymptotes.
Since the function is always positive, there are no \(x\)-intercepts. There is a \(y\)-intercept at \(y=e^{0}=1\).
We first take the limit as \(x\) goes to \(\infty\). \[ \lim_{x \to \infty} e^{-x^2} = \lim_{x \to \infty} \dfrac{1}{e^{x^2}} = 0 \] Then as \(x\) goes to \(-\infty\). \[ \lim_{x \to -\infty} e^{-x^2} = \lim_{x \to -\infty} \dfrac{1}{e^{x^2}} = 0 \] So there is a horizantal asymptote at \(x=0\).
The function is never undefined. So there are no vertical asymptotes.

Next we take the first derivative of \(e^{-x^2}\): \[f'=e^{-x^2}(-2x)\] and find the critical points. \[\begin{aligned} e^{-x^2}(-2x)&=0 \\ x&=0 \end{aligned}\] So there is a critical point at \(x=0\). We next test intervals on either side of the critical point to see if \(x=0\) is a maximum, minimum or inflection point. \[\begin{aligned} f'(-1)=e^{-(-1)^2}(-2(-1)) &=2e^{-1} \gt 0\\ f'(1)=e^{-(1)^2}(-2(1)) &= -2e^{-1} \lt 0 \end{aligned}\] So the function is increasing on \((-\infty, 0)\) and decreasing on \((0,\infty)\). So \(x=0\) is a maximum. We next find the function value at \(x=0\). \[ f(0)=e^{-(0)^2}=1 \] We plot this information on a numberline.

We next look at the second derivative and find secondary critcal points. \[\begin{aligned} f''(x)&=e^{-x^2}(4x^2)+e^{-x^2}(-2) \\ 0&=e^{-x^2}(4x^2)-2e^{-x^2} \\ 2e^{-x^2}&=e^{-x^2}(4x^2) \\ 2&=4x^2 \\ x&=\pm\dfrac{1}{\sqrt2} \end{aligned}\] So \(x=\dfrac{1}{\sqrt2}\) and \(x=-\dfrac{1}{\sqrt2}\) are secondary critcal points. To find the concavity, we look at values of \(f''\) in the intervals on either side of our secondary critical points, by testing values at a point in each interval. \[\begin{aligned} f''(-1)&=e^{-(-1)^2}(4(-1)^2)+e^{-(-1)^2}(-2) = 2e^{-1} \gt 0 \\ f''(0)&=e^{-(0)^2}(4(0)^2)+e^{-(0)^2}(-2) = -2 \lt 0\\ f''(1)&=e^{-(1)^2}(4(1)^2)+e^{-(1)^2}(-2) = 2e^{-1} \gt 0 \end{aligned}\] So \(e^{-x^2}\) is concave up on \(\left(-\infty,-\dfrac{1}{\sqrt2}\right)\) and \(\left(\dfrac{1}{\sqrt2},\infty\right)\) and concave down on \(\left(-\dfrac{1}{\sqrt2},\dfrac{1}{\sqrt2}\right)\).

Next we find the function values at our secondary critical points. \[\begin{aligned} f\left(\dfrac{1}{\sqrt2}\right) &=e^{-(1/\sqrt{2})^2} = e^{-1/2} = \dfrac{1}{\sqrt{e}}\\ f\left(-\dfrac{1}{\sqrt2}\right) &=e^{-(-1/\sqrt{2})^2} = e^{-1/2} = \dfrac{1}{\sqrt{e}} \end{aligned}\]

We sumarize this information on a numberline:

To sketch the graph, we first plot the \(y\)-intercept, the critical point and the secondary critical points, along with what we know about slope and concavity at these points:

Finally, to get the final graph, we connect the dots.

mjs
Sketch the graph of the function \(g(x)=\dfrac{x}{1+x^2}\).

Use information from the function, the first, and the second derivative to get reference points to sketch the graph.

The Sketch:

We first find the \(x\)-intercept: \[ g(x)=\dfrac{x}{1+x^2}=0 \quad \text{when} \quad x=0 \] Next we find the \(y\)-intercept: \[ y=g(0)=\dfrac{0}{1+0^2}=0 \] So the only \(x\) and \(y\)-intercept is at \((0,0)\). Next we check for horizantal asymptotes.
At the right, since \(x \gt 0\), \(g(x)=\dfrac{x}{1+x^2} \gt 0\) and: \[ \lim_{x \to \infty}\dfrac{x}{1+x^2} =\lim_{x \to \infty}\dfrac{\dfrac{1}{x}}{\dfrac{1}{x^2}+1} = \dfrac{0}{1} = 0 \] At the left, since \(x \lt 0\), \(g(x)=\dfrac{x}{1+x^2} \lt 0\) and: \[ \lim_{x \to -\infty}\dfrac{x}{1+x^2} =\lim_{x \to -\infty}\dfrac{\dfrac{1}{x}}{\dfrac{1}{x^2}+1} = \dfrac{0}{1} = 0 \] So the horizontal asymptotes at both \(\pm\infty\) are \(y=0\) but \(g(x)\) approaches \(0\) from above on the right and from below on the left.
We plot this information on a number line:

Next we find the first derivative and the critical points: \[ g'(x)=\dfrac{(1+x^2)-x(2x)}{(1+x^2)^2} =\dfrac{1-x^2}{(1+x^2)^2} \] \[ g'(x)=0 \quad \text{when} \quad 1-x^2=0 \quad \text{or} \quad x=\pm1 \] The function values at the critical points are: \[\begin{aligned} g(-1)&=\dfrac{(-1)}{1+(-1)^2}&&=-\,\dfrac{1}{2} \\ g(1)&=\dfrac{(1)}{1+(1)^2}&&=\dfrac{1}{2} \end{aligned}\] So, the critical points are \((-1,-\dfrac{1}{2})\) and \((1,\dfrac{1}{2})\). To find where the function is increasing and decreasing, we test values between the critical points: \[\begin{aligned} g'(-2)&=\dfrac{1-(-2)^2}{(1+(-2)^2)^2} &&=\dfrac{-3}{25} \lt 0 \\ g'(0)&=\dfrac{1-(0)^2}{(1+(0)^2)^2} &&= 1 \gt 0 \\ g'(2)&=\dfrac{1-(2)^2}{(1+(2)^2)^2} &&= \dfrac{-3}{25} \lt 0 \\ \end{aligned}\] So the function in decreasing on \((-\infty,-1)\) and \((1,\infty)\) and increasing on \((-1,1)\). Consequently, there is a minimum at \((-1,-\dfrac{1}{2})\) and a maximum at \((1,\dfrac{1}{2})\). We add this information to the number line:

Next, we find the second derivative and the secondary critical points: \[\begin{aligned} g''&=\dfrac{(1+x^2)^2[-2x]-(1-x^2)[2(1+x^2)(2x)]}{(1+x^2)^4}\\ &=\dfrac{-2x-2x^3-4x+4x^3}{(1+x^2)^3} =\dfrac{2x^3-6x}{(1+x^2)^3} \end{aligned}\] \[\begin{aligned} g''(x)=0 \quad &\text{when} \quad 2x^3-6x=2x(x^2-3)=0 \\ \quad &\text{or} \quad x=0,\pm\sqrt{3} \end{aligned}\] The function values at the secondary critical points are: \[\begin{aligned} g(-\sqrt{3})&=\dfrac{(-\sqrt{3})}{1+(-\sqrt{3})^2}&&=\dfrac{-\sqrt{3}}{4} \approx -0.433 \\ g(0)&=\dfrac{(0)}{1+(0)^2}&&=0 \\ g(\sqrt{3})&=\dfrac{(\sqrt{3})}{1+(\sqrt{3})^2}&&=\dfrac{\sqrt{3}}{4} \approx 0.433 \end{aligned}\] So, the secondary critical points are \((-\,\sqrt{3},-\,\dfrac{\sqrt{3}}{4})\), \((0,0)\) and \((\sqrt{3},\dfrac{\sqrt{3}}{4})\). To find where the function is concave up and down, we test values between the secondary critical points: \[\begin{aligned} g''(-2)&=\dfrac{2(-2)^3-6(-2)}{(1+(-2)^2)^3} &&= \dfrac{-4}{125} \lt 0 \\ g''(-1)&=\dfrac{2(-1)^3-6(-1)}{(1+(-1)^2)^3} &&= \dfrac{1}{2} \gt 0 \\ g''(1)&=\dfrac{2(1)^3-6(1)}{(1+(1)^2)^3} &&= \dfrac{-1}{2} \lt 0 \\ g''(2)&=\dfrac{2(2)^3-6(2)}{(1+(2)^2)^3} &&= \dfrac{4}{125} \gt 0 \\ \end{aligned}\] The function is concave down on \((-\infty,-\,\sqrt{3})\) and \((0,\sqrt{3})\) and concave up on \((-\,\sqrt{3},0)\) and \((\sqrt{3},\infty)\). So there are inflection points at all three secondary critical points. We add this information to the number line.

Finally, we plot all the information in a graph, and connect the dots.

f. Interpreting Graphs

Do the Tutorial until you feel confident.