10. Derivatives and Tangent Lines

c. Derivative as a Function

2. Sloe, Increasing and Decreasing

All the Applications of Derivatives at a Point, can now be done using functions. And there are several additional applications, like acceleration and jerk on the next page.

Slope

The derivative of the function, \(f(x)\), is the function, \(f'(x)\), which gives the slope of the tangent line at each point.

Find the slope of the tangent line to a function, \(f(x)=x^3\), at each of the points \(x=-1\), \(x=1\), \(x=2\) and \(x=3\).

\(f'(x)=3x^2\)
\(f'(-1)=3 \qquad f'(1)=3 \qquad f'(2)=12 \qquad f'(3)=27\)

Rather than compute the \(3\) slopes separately, we compute them all at once: \[\begin{aligned} f'(x)&=\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h} \\ &=\lim_{h\to0} \dfrac{(x+h)^3-x^3}{h} \\ &=\lim_{h\to0} \dfrac{(x^3+3x^2h+3xh^2+h^3)-x^3}{h} \\ &=\lim_{h\to0} \dfrac{3x^2h+3xh^2+h^3}{h} \\ &=\lim_{h\to0} (3x^2+3xh+h^2)=3x^2 \end{aligned}\]

Then \[ f'(-1)=3 \qquad f'(1)=3 \qquad f'(2)=12 \qquad f'(3)=27 \]

Increasing and Decreasing

Since the first derivative of \(f(x)\) is the slope of the tangent line, we can use it to tell if the function is increasing or decreasing.

In a region where the function is getting bigger as we move from left to right, the curve is increasing. If the derivative is positive, \(f'(x)\gt0\), then the slope is positive and the function is increasing.

The function at the right is increasing on \([-2,0]\) and \([2,\infty)\) as indicated by the red tangent lines.

The graph is a W shape, highlighting that it is increasing from -2 to
the origin and from 2 to infinity.

In a region where the function is getting smaller as we move from left to right, the curve is decreasing. If the derivative is negative, \(f'(x)\lt0\), then the slope is negative and the function is decreasing.

The function at the right is decreasing on \((-\infty,-2]\) and \([0,2]\) as indicated by the red tangent lines.

The graph is the same W shape, highlighting that it is decreasing from
- infinity to -2 and from the origin to 2.

Find the intervals where the graph of \(f(x)=x^4-8x^2\) is increasing and decreasing.

We find the derivative and factor it: \[\begin{aligned} f'(x)&=\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h} \\ &=\lim_{h\to0} \dfrac{[(x+h)^4-8(x+h)^2]-(x^4-8x^2)}{h} \\ &=\lim_{h\to0} \dfrac{1}{h}\Big[(x^4+4x^3h+6x^2h^2+4xh^3+h^4)\Big. \\ &\qquad\qquad\quad\Big.-8(x^2+2xh+h^2)-(x^4-8x^2)\Big] \\ &=\lim_{h\to0} \dfrac{4x^3h+6x^2h^2+4xh^3+h^4-16xh-8h^2}{h} \\ &=\lim_{h\to0} (4x^3+6x^2h+4xh^2+h^3-16x-8h) \\ &=4x^3-16x =4x(x-2)(x+2) \end{aligned}\] The interesting points are where \(f'(x)=0\), i.e. \(x=0,2,-2\). These are called the critical points of the function. We draw a number line and mark in the critical points: A horizontal number line labeled from −5 to 5 has solid dots at −2, 0
and 2. Above each dot is a 0, indicating the derivative equals zero at those
three x values. We now test a point in each interval to see if \(f'(x)=4x(x-2)(x+2)\) is positive or negative. \[\begin{aligned} f'(-3)&=4(-3)(-5)(-1)&&=-60 \lt 0 \\ f'(-1)&=4(-1)(-3)(1) &&=12 \gt 0 \\ f'(1)&=4(1)(-1)(3) &&=-12 \lt 0 \\ f'(3)&=4(3)(1)(5) &&=60 \gt 0 \end{aligned}\] We have marked these on the number line along with a line to show increasing or decreasing:

So \(f(x)\) is decreasing on \((-\infty,-2]\cup[0,2]\) and is increasing on \([-2,0]\cup[2,\infty)\).

These facts are verified in a plot.

The graph shows a W shaped wiggle, symmetric about the y-axis. It
rises on the far left, dips to a low point at x = -2, rises to a peak at
the origin, dips again to a matching low point at x = 2, then rises upward
on the far right. Three red dots mark these turning points: with local
minima at x = -2 and 2 and a local maximum at the origin.

Find the intervals where the graph of \(f(x)=3x^4-16x^3+24x^2\) is increasing and decreasing.

Factor the derivative and determine where it is positive and negative.

\(f(x)\) is decreasing on \((-\infty,0]\) and is increasing on \([0,\infty)\).

We find the derivative and factor it: \[\begin{aligned} f'(x)&=\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h} \\ &=\lim_{h\to0} \dfrac{1}{h}\left(\rule{0pt}{12pt}\left[\rule{0pt}{9pt}3(x+h)^4-16(x+h)^3+24(x+h)^2\right]\right. \\ &\qquad\qquad\quad-\left.\rule{0pt}{12pt}\left[\rule{0pt}{9pt}3x^4-16x^3+24x^2\right]\right) \\ &=\lim_{h\to0} \dfrac{1}{h}\left(\rule{0pt}{12pt}\left[\rule{0pt}{9pt}3(x^4+4x^3h+O(h^2)) -16(x^3+3x^2h+O(h^2))\right.\right. \\ &\qquad\qquad\quad\left.\rule{0pt}{12pt}\left.\rule{0pt}{9pt}+24(x^2+2xh+h^2)\right] -\left[\rule{0pt}{9pt}3x^4-16x^3+24x^2\right]\right) \\ &=\lim_{h\to0} \dfrac{1}{h}\left(12x^3h-48x^2h+48xh+O(h^2)\right) \\ &=\lim_{h\to0} \left(12x^3-48x^2+48x+O(h)\right) \\ &=12x^3-48x^2+48x=12x(x^2-4x+4)=12x(x-2)^2 \end{aligned}\] Where \(O(h^n)\) means terms with powers of \(h\) of degree \(n\) or higher. The interesting points are where \(f'(x)=0\), i.e. \(x=0,2\). We draw a number line and mark in these critical points: The plot shows a number line labeled from -5 to 5 with solid dots
at x = 0 and x = 2. Above each dot is a 0 to indicate the derivative is 0. We now test a point in each interval to see if \(f'(x)=12x(x-2)^2\) is positive or negative. \[\begin{aligned} f'(-1)&=12(-1)(-3)^2 &&=-108 \lt 0 \\ f'(1)&=12(1)(-1)^2 &&=12 \gt 0 \\ f'(3)&=12(3)(1)^2 &&=36 \gt 0 \end{aligned}\] We have marked these on the number line along with a line to show increasing or decreasing:

Notice \(f(x)\) is decreasing on \((-\infty,0]\) and is increasing on \([0,\infty)\), even though it becomes horizontal at \(x=2\).