10. Derivatives and Tangent Lines

c. Derivative as a Function

2. Slope, 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.

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.

---

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

We find the derivative and factor it: \[\begin{aligned} f'(x)&=\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h} \\ &=\lim_{h\to0} \dfrac{[(x+h)^4-8(x+h)^2]-(x^4-8x^2)}{h} \\ &=\lim_{h\to0} \dfrac{1}{h}\left[(x^4+4x^3h+6x^2h^2+4xh^3+h^4)\right. \\ &\qquad\qquad\quad\left.-8(x^2+2xh+h^2)-(x^4-8x^2)\right] \\ &=\lim_{h\to0} \dfrac{4x^3h+6x^2h^2+4xh^3+h^4-16xh-8h^2}{h} \\ &=\lim_{h\to0} (4x^3+6x^2h+4xh^2+h^3-16x-8h) \\ &=4x^3-16x =4x(x-2)(x+2) \end{aligned}\] 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: \(\begin{array}{r} f'(x)= \\ x= \end{array}\) We now test a point in each interval to see if \(f'(x)=4x(x-2)(x+2)\) is positive or negative. \[\begin{aligned} f'(-3)&=4(-3)(-5)(-1)=-60 \\ f'(-1)&=4(-1)(-3)(1)=12 \\ f'(1)&=4(1)(-1)(3)=-12 \\ f'(3)&=4(3)(1)(5)=60 \\ \end{aligned}\] We have marked these on the number line: \(\begin{array}{r} f'(x)= \\ x= \end{array}\)

So \(f(x)\) is decreasing on \((-\infty,-2]\cup[0,2]\) and is increasing on \([-2,0]\cup[2,\infty)\). These are marked on the number line: