# Monotonic Functions (Increasing and Decreasing)

Monotonic simply means either increasing or decreasing. A function is monotonic if one or the other holds. In this article, I discuss at length finding extrema of monotonic functions. An essential tool in this regard is the First Derivative Test. Finding extrema becomes much easier when you have mastered this theorem.

All relative extrema of a functions’ graph must occur at critical numbers of the function (where the derivative is undefined or zero). We discuss the first derivative test and illustrate its use with several examples. Exercises are given at the end.

## Increasing and Decreasing Functions

To determine where a function $f$ is increasing or decreasing, we begin by finding the critical numbers. These numbers divide the $x$-axis into intervals, and we test the sign of $f'(x)$ in each of these intervals. This procedure is often called the first derivative test and can be used to determine local extrema and intervals of monotonicity.

Definition. A function $f$ is called increasing on an interval $I$ if $f\left(x_1\right) < f\left(x_2\right)$ whenever $x_1<x_2$ on $I.$ A function $f$ is called decreasing on an interval $I$ if $f\left(x_1\right) > f\left(x_2\right)$ whenever $x_1>x_2$ on $I.$

Definition. A function $f$ is called monotonic on an interval $I$ if it is either increasing or decreasing on $I.$

Theorem. Suppose $f$ is continuous on $[a,b]$ and differentiable on $(a,b).$

(1) If $f'(x)>0$ for all $x$ in $(a,b),$ then $f$ is increasing on $(a,b).$

(2) If $f'(x)<0$ for all $x$ in $(a,b),$ then $f$ is decreasing on $(a,b).$

Example. Find where the function $$f(x)=3x^4-4x^3-12x^2+5$$ is increasing and decreasing; that is determine the intervals where $f$ is monotonic. A sketch of the function $f(x)=3x^4-4x^3-12x^2+5.$

Solution. The derivative of $f$ is \begin{equation} f'(x)=12x^3-12x^2-24x=12x(x-2)(x+1). \end{equation} Since $f$ is continuous and differentiable, to test where the function is monotonic we divide the $x$-axis according to the sign of $f'(x)$ which depending on the signs of $12x,$ $x-2,$ and $x+1.$ We put our results into the following table: \begin{equation} \begin{array}{c|c|c|c|c|l} \text{Interval} & 12x & x-2 & x+1 & f'(x)\text{ } & f \\ \hline x<-1 & – & – & – & – & \text{decreasing on } (-\infty ,-1) \\ -12 & + & + & + & + & \text{increasing on } (2,\infty ) \end{array} \end{equation} as needed.

## The First Derivative Test

Theorem. (First Derivative Test) Suppose that $c$ is a critical number of a function that is continuous on $[a,b].$ Then the following statements hold:

(1) If $f'(x)>0$ on $(a,c)$ and $f'(x)<0$ on $(c,b),$ then $f$ has a relative (local) maximum at $c.$

(2) If $f'(x)<0$ on $(a,c)$ and $f'(x)>0$ on $(c,b),$ then $f$ has a relative (local) maximum at $c.$

(3) If neither hold then $f$ has no relative (local) extremum at $c.$

Example. Apply the First Derivative Test to find the local extrema of the function $$f(x)=x(1-x)^{2/5}$$ and sketch its graph. A sketch of the function $f(x)=x(1-x)^{2/5}.$

Solution. First we find the critical numbers of $f$ by solving $f'(x)=0$ and determining where $f'(x)$ is undefined but $f(x)$ is defined. The derivative of $f$ is \begin{align*} f'(x) & =(1-x)^{2/5}+\frac{2x}{5}(1-x)^{-3/5}(-1) \\ & =\frac{5(1-x)-2x}{5(1-x)^{3/5}} \\ & =\frac{5-7x}{5(1-x)^{3/5}}. \end{align*} Solving $f'(x)=0$ we find $x=5/7.$ Also $f'(1)$ does not exist but $f(1)=0;$ and therefore the only critical numbers of $f$ are $x=5/7$ and $x=0.$ We determine the local extrema using the following table \begin{align*} \begin{array}{c|c|c|c|l} \text{Interval} & 5-7x & (1-x)^{3/5} & f'(x) & f(x) \\ \hline x<\frac{5}{7} & + & + & + & \text{ increasing on} \left(-\infty ,\frac{5}{7}\right) \\ \frac{5}{7}1 & – & – & + & \text{ increasing on }(1,+\infty ) \end{array} \end{align*} Therefore, $\left(\frac{5}{7},f\left(\frac{5}{7}\right)\right)$ is a local maximum and $(1,f(1))$ is a local minimum. Here is the graph of the function $f(x)=x(1-x)^{2/5}.$ Notice there is a corner at $(1,f(1))$ because $f$ is defined there but $f’$ is not.

## Finding Extrema

Example. Find the local and absolute extrema values of the function $$f(x)=x^3(x-2)^2$$ on the interval $-1\leq x\leq 3.$ Sketch the graph. A sketch of the function $f(x)=x^3(x-2)^2.$

Solution. First we notice that $f$ is a polynomial and so is continuous and differentiable for all real numbers. Next we find the critical numbers of $f$ by solving $f'(x)=0$ and determining where $f'(x)$ is undefined, but $f(x)$ is defined. The derivative of $f$ is \begin{align*} f'(x) & =3x^2(x-2)^2+2x^3(x-2) \\ & =x^2(x-2)(5x-6). \end{align*} To find the critical numbers we set $f'(x)=0$ and obtain $x=0,2,6/5.$ We determine the local extrema and absolute extrema using the following table: \begin{align*} \begin{array}{c|c|c|c|c|l} \text{Interval} & x^2 & x-2 & 5x-6 & f'(x)\text{ } & f \\ \hline -1<x<0 & + & – & – & + & \text{ increasing on} (-1,0) \\ 0<x<\frac{6}{5} & + & – & – & + & \text{ increasing on } \left(0, \frac{6}{5}\right) \\ \frac{6}{5}<x<2 & + & – & + & – & \text{ decreasing on }\left(\frac{6}{5},2\right) \\ 2<x<3 & + & + & + & + & \text{ increasing on } (2,3) \end{array} \end{align*} The function $f$ does not have a local extrema at $x=0.$ The local maximum is $\left(\frac{6}{5},f\left(\frac{6}{5}\right)\right)$ and the local minimum is $(2,f(2)).$ Since $f$ is a continuous function we can use the extreme value theorem to determine absolute extrema. We compute the functional values at the endpoints, namely $f(-1)=-9$ and $f(3)=27.$ Therefore, the absolute maximum is $f(3)=27$ and the absolute minimum is $f(-1)=-9$.

## Exercises on Monotonic Functions

Exercise. For each of the following functions determine the critical points and apply the first derivative test to determine the intervals where the function is increasing or decreasing, and all local extrema.

$(1) \quad \displaystyle f(x)=(x-1)^2(x+2)$

$(2) \quad \displaystyle f(x)=(x-1)e^{-x}$

$(3) \quad \displaystyle f(x)=x^{-1/3}(x+2)$

$(4) \quad \displaystyle f(\theta )=3\theta ^2-4\theta ^3$

$(5) \quad \displaystyle h(r)=(r+7)^3$

$(6) \quad \displaystyle g(x)=x^2\sqrt{5-x}$

$(7) \quad \displaystyle f(x)=\frac{x^3}{3x^2+1}$

$(8) \quad \displaystyle h(x)=x^{1/3}\left(x^2-4\right)$

$(9) \quad \displaystyle f(x)=e^{2x}+e^{-x}$

$(10) \quad \displaystyle f(x)=x \ln x$

$(11) \quad \displaystyle f(x)=(x+1)^2$

$(12) \quad \displaystyle f(x)=-x^2-6x-9$

$(13) \quad \displaystyle k(x)=x^3+3x^2+3x+1$

Exercise. Sketch the graph of a differentiable function $y=f(x)$ through the point $(1,1)$ that satisfies the following conditions:

$(1) \quad f'(1)=0$

$(2) \quad f'(x)>0$ for $x<1$ \item $f'(x)<0$ for $x>1$

$(3) \quad f'(x)>0$ for $x\neq 1$

$(4) \quad f'(x)<0$ for $x\neq 1$

Exercise. Sketch the graph of a differentiable function $y=f(x)$ that satisfies the following condition.

(1) a local minimum at $(1,1)$

(2) a local maximum at $(3,3)$

(3) local maximum at $(1,1)$

(4) a local minimum at $(3,3)$

Exercise. Sketch the graph of a differentiable function $y=h(x)$ that satisfies all of the following conditions. $h(0)=0,$ for all $x$, $-2\leq h(x)\leq 2,$ $h'(x)\to +\infty$ as $x\to 0^-,$ and $h'(x)\to +\infty$ as $x\to 0^+.$ David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.