Short and Sweet Calculus

## 4.1Increasing and Decreasing Functions

• A function is increasing if as we move along the curve from left to right the curve is rising (Figure [fig:Ch6-Increasing-Fig1](a)). Mathematically speaking, a function $$f$$ is increasing on an interval $$I$$ if $f(x_{1})<f(x_{2})\quad\text{whenever}\quad x_{1}<x_{2}\text{ in }I$

• A function is decreasing if as we move along the curve from left to right, the curve is falling (Figure [fig:Ch6-Increasing-Fig1](b)). Mathemtically speaking, $$f$$ is a decreasing function on $$I$$ if $f(x_{1})>f(x_{2})\quad\text{whenever}\quad x_{1}<x_{2}\text{ in }I.$

• A function is constant if its graph is horizontal (Figure [fig:Ch6-Increasing-Fig1](c)). Mathematically speaking, we say $$f$$ is constant on $$I$$ if $f(x_{1})=f(x_{2})\quad\text{for all points }x_{1}\text{ and }x_{2}\text{ in }I.$

 (a) Increasing function. $$x_{1} A function may be sometimes increasing and sometimes decreasing (or neither). For example, the graph of \(f(x)=2x^{3}-12x^{2}+18x-2$$ is illustrated in Figure 4.1. At $$x=1$$ the function ceases to increase and begins decreasing; at $$x=3$$, the reverse is true. At $$x=1$$ and $$x=3$$ the tangent to the curve is evidently parallel to the $$x$$-axis, meaning the slope is zero. We will talk about this kind of points in Section [sec:Ch4-Extreme-values]. Graph of $$f(x)=x^{3}-12x^{2}+18x-2$$.

#### Slope of graph and increasing and decreasing functions

It is evident from Figure 4.2 that on the interval where a function $$f(x)$$ is increasing, the tangent line makes a positive acute angle $$\theta$$ with the positive $$x$$-axis; hence $\text{slope}=\tan\theta=f'(x)>0$ and on the interval where the function is decreasing, the tangent line makes an obtuse angle $$\theta$$ with the positive $$x$$-axis; hence $\text{slope}=\tan\theta=f'(x)<0.$

#### 4.1.0.1 Tests for determining where a function is increasing and where decreasing

From the above observations, we may guess that $$f$$ is increasing function if $$f'(x)>0$$ and is decreasing if $$f'(x)<0$$. Let’s see how we can verify that.

First let us suppose that $$f'(x)=dy/dx$$ is positive. Then since $$dy/dx$$ is the limit of $$\Delta y/\Delta x$$, it follows that, we have $\frac{\Delta y}{\Delta x}>0,$ for sufficiently small values of $$\Delta x$$. This means that if $$\Delta x=x-x_{0}$$ is positive then $$\Delta y=f(x)-f(x_{0})$$ is also positive; that is, $\Delta x=x-x_{0}>0\Rightarrow\Delta y=f(x)-f(x_{0})>0$ or $x_{0}<x\Rightarrow f(x_{0})<f(x).$ Therefore, an increase of $$x$$ causes an increase of $$y=f(x)$$. From this argument, we learn that we can verify a differentiable function is increasing where its derivative is positive $f'(x)=\frac{dy}{dx}=\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}>0.$ Analogously, if $$dy/dx<0$$, it follows that $$\Delta y/\Delta x<0$$ for sufficiently small values of $$\Delta x$$; that is if $$\Delta x>0$$, then $$\Delta y<0$$ or an increase of $$x$$ causes a decrease of $$y=f(x)$$.

4.1. Increasing/Decreasing Test.

(a) If $$f'(x)>0$$ for every $$x$$ in an interval, then $$f$$ is increasing $$(\nearrow)$$ on that interval

(b) If $$f'(x)<0$$ for every $$x$$ in an interval, then $$f$$ is decreasing $$(\searrow)$$ on that interval.

(c) If $$f'(x)=0$$ for every $$x$$ in an interval, then $$f$$ is constant on that interval.

In applying this theorem, we need to determine the sign of $$f'(x)$$. If $$f'(x)$$ is a polynomial, this may be conveniently done by breaking it up into factors and determining the sign of each factor. A factor of the form $$x-a$$ is negative when $$x<a$$ and is positive when $$x>a$$.

Suppose, then, we wish to determine the sign of $(x+2)(x-1)(x-3).$

1. When $$-\infty<x<-2$$, all three factors are negative, and thus the product is negative $(-)(-)(-)=(-).$

2. When $$-2<x<1$$, the first factor is positive, and the other factors are negative. Therefore, the product is positive $(+)(-)(-)=(+).$

3. When $$1<x<3$$, the first two factors are positive, but the last factor is negative. Therefore, the product is negative $(+)(+)(-)=(-).$

4. When $$3<x<+\infty$$, all three factors are positive, and the product is positive too $(+)(+)(+)=(+).$

Obviously the product is zero at $$-2,1$$, and $$3$$. We may summarize these into the following “sign table.”

$$x$$ $$-\infty$$ $$-2$$ 1 3 $$+\infty$$
sign of $$(x+2)$$ $$-$$ $$0$$ $$+$$ $$+$$ $$+$$ $$+$$ $$+$$
sign of $$(x-1)$$ $$-$$ $$-$$ $$-$$ $$0$$ $$+$$ $$+$$ $$+$$
sign of $$(x-3)$$ $$-$$ $$-$$ $$-$$ $$-$$ $$-$$ $$0$$ $$+$$
$$\therefore$$ sign of product $$-$$ $$0$$ $$+$$ $$0$$ $$-$$ $$0$$ $$+$$

Example 4.1. Determine where $$f(x)=x^{2}e^{-x}$$ is increasing and where it is decreasing.

Solution

Let’s find the derivative of $$f$$: $f(x)=x^{2}e^{-x}\Rightarrow f'(x)=2xe^{-x}-x^{2}e^{-x}=e^{-x}(2x-x^{2}).$ Because $$e^{-x}>0$$, the sign of $$f'(x)$$ is solely determined by $$2x-x^{2}=x(2-x).$$

$$x$$ $$-\infty$$ 0 2 $$+\infty$$
sign of $$x$$ $$—$$ $$0$$ $$+++$$ $$+$$ $$+++$$
sign of $$2-x$$ $$+++$$ $$+$$ $$+++$$ $$0$$ $$—$$
$$\therefore$$ sign of $$f'(x)$$ $$—$$ $$0$$ $$+++$$ $$0$$ $$—$$
Increaing/Decreasing $$f(x)$$ $$\searrow$$ $$\nearrow$$ $$\searrow$$

This table shows that $$f(x)$$ is decreasing on $$(-\infty,0]\cup[2,+\infty)$$ (because $$f'(x)<0$$) and is increasing on $$[0,2]$$ (because $$f'(x)>0$$). The graph of $$f$$ is shown in the following figure.