19. Properties of Graphs

e. Curve Sketching

We now put everything together to get an idea about how to roughly graph a function. Some information will be redundant. It is best to learn by example.

For the function \(f(x)=\dfrac{1}{6}x^6-\dfrac{2}{5}x^5+\dfrac{1}{4}x^4\), find the intercepts, asymptotes, intervals where the function is increasing, decreasing, concave up and concave down, local minima, local maxima and inflection points. Then plot it.

• \(y\)-intercept: The \(y\)-intercept is at \(f(0)=0\).
• \(x\)-intercepts: To find the \(x\)-intercepts, we try factoring the function. We first write: \[ f(x)=\dfrac{1}{60}x^4(10x^2-24x+15) \] To find the zeros of the second factor, we use the quadratic formula: \[ x=\dfrac{24\pm\sqrt{4\cdot144-4\cdot150}}{20} =\dfrac{12\pm i\sqrt{6}}{10} \] Since the solutions are complex, there are no real solutions to the quadratic factor and the the only \(x\)-intercept is \(x=0\). Further, since \(f(-1)=\dfrac{49}{60}\) and \(f(1)=\dfrac{1}{60}\), the function is always positive except at \(x=0\).
• vertical asymptotes: The function is defined for all \(x\). So there are no vertical asymptotes.
• horizontal and slant asymptotes: Since the function looks like \(\dfrac{1}{6}x^6\) for large \(x\), there are no horizontal or slant asymptotes.
• increasing and decreasing: The derivative is: \[ f'(x)=x^5-2x^4+x^3=x^3(x^2-2x+1)=x^3(x-1)^2 \] So the critical points are at \(x=0\) and \(x=1\). The value at the new point is: \[ f(1)=\dfrac{1}{6}-\dfrac{2}{5}+\dfrac{1}{4}=\dfrac{1}{60} \] We plot these on a number line. Then we evaluate the derivative at a point in each interval to see if it is positive or negative. \[\begin{aligned} f'(-1)&=-4 \lt 0 \\ f'(.5)&=.125(.5)^2 \gt 0 \\ f'(2)&=8 \gt 0 \end{aligned}\] So the function is decreasing on \((-\infty,0)\) and increasing on \((0,\infty)\). Notice the derivative is \(0\) at \(x=1\) but the function continues to increase on both sides.
• concavity and inflection points: The second derivative is: \[ f''(x)=5x^4-8x^3+3x^2=x^2(5x^2-8x+3)=x^2(5x-3)(x-1) \] So the secondary critical points are \(x=0\), \(x=\dfrac{3}{5}\) and \(x=1\). The value at the new point is: \[ f\left(\dfrac{3}{5}\right) =\dfrac{1}{6}\left(\dfrac{3}{5}\right)^6 -\dfrac{2}{5}\left(\dfrac{3}{5}\right)^5 +\dfrac{1}{4}\left(\dfrac{3}{5}\right)^4 =\dfrac{567}{62500} \approx0.009 \] We plot these on a number line. Then we evaluate the second derivative at a point in each interval to see if it is positive or negative. \[\begin{aligned} f''(-1)&=(1)(-8)(-2)=16 \gt 0& f''(.5)&=(.25)(-.5)(-.5) \gt 0 \\ f''(.75)&=.75^2(.75)(-.25) \lt 0& f''(2)&=4(7)(1)=28 \gt 0 \end{aligned}\] So the function is concave up on \((-\infty,\dfrac{3}{5})\), concave down on \((\dfrac{3}{5},1)\) and concave up again on \((1,\infty)\). Notice the second derivative is \(0\) at \(x=0\) but the function continues to be concave up on both sides. So there are inflection points only at \(x=\dfrac{3}{5}\) and \(x=1\).
• local minima, local maxima and horizontal inflection points: We evaluate the second derivative at each critical point, \(x=0\) and \(x=1\): \[ f''(0)=0^2(0-3)(0-1)=0 \qquad f''(1)=1(5-3)(0)=0 \] So the Second Derivative Test fails at both critical points. We turn to the First Derivative Test. The function decreases on \((-\infty,0)\) and increases on \((0,1)\) and \((1,\infty)\). So there is a local minimum at \(x=0\) and a horizontal inflection point at \(x=1\). There is no local maximum.
• plot: (Click on each button.)

For the function \(f(x)=\dfrac{x^2+4}{x^2-4}\), find the intercepts, asymptotes, intervals where the function is increasing, decreasing, concave up and concave down, local minima, local maxima and inflection points. Then plot it.

• \(y\)-intercept: The \(y\)-intercept is at \(f(0)=\dfrac{+4}{-4}=-1\).
• \(x\)-intercepts: Since the numerator, \(x^2+4\), is never \(0\), there are no \(x\)-intercepts.
• vertical asymptotes: The denominator, \(x^2-4\), is \(0\) at \(x=\pm2\). So these are the vertical asymptotes. We plot the location of the intercept and vertical asymptotes: We rewrite the function as \[ f(x)=\dfrac{x^2+4}{(x+2)(x-2)} \] and compute four limits to determine how the curve approaches the vertical asymptotes: \[\begin{aligned} \lim_{x\to-2^-}\dfrac{x^2+4}{(x+2)(x-2)} &=``\dfrac{4+4}{(-2^-+2)(-2^--2)}"=``\dfrac{8}{(0^-)(-4)}"=\infty \\ \lim_{x\to-2^+}\dfrac{x^2+4}{(x+2)(x-2)} &=``\dfrac{4+4}{(-2^++2)(-2^+-2)}"=``\dfrac{8}{(0^+)(-4)}"=-\infty \\ \lim_{x\to2^-}\dfrac{x^2+4}{(x+2)(x-2)} &=``\dfrac{4+4}{(2^-+2)(2^--2)}"=``\dfrac{8}{(4)(0^-)}"=-\infty \\ \lim_{x\to2^+}\dfrac{x^2+4}{(x+2)(x-2)} &=``\dfrac{4+4}{(2^++2)(2^+-2)}"=``\dfrac{8}{(4)(0^+)}"=\infty \end{aligned}\] So the asymptote at \(x=-2\) is approached up-down while the asymptote at \(x=2\) is approached down-up.
• horizontal and slant asymptotes: We take the limits as \(x\to\pm\infty\): \[\begin{aligned} \lim_{x\to-\infty}\dfrac{x^2+4}{x^2-4}=1 \\ \lim_{x\to\infty}\dfrac{x^2+4}{x^2-4}=1 \end{aligned}\] So the horizontal asymptote is \(y=1\) at both \(-\infty\) and \(\infty\). To see if these asymptotes are approached from above or below, we look at the sign of \[ f(x)-1=\dfrac{x^2+4}{x^2-4}-1=\dfrac{(x^2+4)-(x^2-4)}{x^2-4} =\dfrac{8}{x^2-4} \] This is positive as \(x\to\pm\infty\). So the asymptotes are approached from above.
• increasing and decreasing: The derivative of \(f(x)=\dfrac{x^2+4}{x^2-4}\) is: \[ f'(x)=\dfrac{(x^2-4)2x-(x^2+4)2x}{(x^2-4)^2}=\dfrac{-16x}{(x^2-4)^2} \] So the critical points are \(x=0\) and \(x=\pm2\). We plot these on a number line. Notice that these happen to be the in the same locations as the intercept and the vertical asymptotes. Next, we evaluate the derivative at a point in each interval to see if it is positive or negative. \[\begin{aligned} f'(-3)&=\dfrac{48}{(9-4)^2} \gt 0& f'(3)&=\dfrac{-48}{(9-4)^2} \lt 0 \\ f'(-1)&=\dfrac{16}{(1-4)^2} \gt 0& f'(1)&=\dfrac{-16}{(1-4)^2} \lt 0 \end{aligned}\] So the function is increasing on \((-\infty,-2)\) and \((-2,0)\) and decreasing on \((0,2)\) and \((2,\infty)\).
• concavity and inflection points: The second derivative of \(f(x)=\dfrac{x^2+4}{x^2-4}\) is: \[\begin{aligned} f''(x)&=\dfrac{(x^2-4)^2(-16)-(-16x)2(x^2-4)2x}{(x^2-4)^4} \\ &=\dfrac{(x^2-4)(-16)-(-16x)4x}{(x^2-4)^3} \\ &=\dfrac{16(3x^2+4)}{(x^2-4)^3} \end{aligned}\] Since the numerator is never \(0\), the only secondary critical points are the vertical asymptotes \(x=\pm2\). We, once again, plot these on a number line. Next, we evaluate the second derivative at a point in each interval to see if it is positive or negative. \[\begin{aligned} f''(-3)&=\dfrac{16(27+4)}{(9-4)^3} \gt 0& f''(3)&=\dfrac{16(27+4)}{(9-4)^3} \gt 0 \\ f''(0)&=\dfrac{16(0+4)}{(0-4)^3} \lt 0 \end{aligned}\] So the function is concave up on \((-\infty,-2)\), concave down on \((-2,2)\) and concave up again on \((2,\infty)\). Notice the second derivative is never \(0\). So there are no inflection points.
• local minima, local maxima and horizontal inflection points: The critical points are \(x=0\) and \(x=\pm2\). However, \(x=\pm2\) are vertical asymptotes. So they cannot be local maxima nor minima. We evaluate the second derivative \(x=0\): \[ f''(0)=\dfrac{16(0+4)}{(0-4)^3}=-1 \lt 0 \] So the Second Derivative Test says\((0,-1)\) is a local minimum. There are no local maxima.
• plot: (Click on each button.)