The First Derivative Test and the Extreme Values of Functions
Definition of Absolute and Relative Extrema
A function $f$ has an absolute minimum at $c$ if $f(c) \leq f(x)$ for all $x$ in the domain of $f$. A function $f$ has an absolute maximum at $c$ if $f(c) \geq f(x)$ for all $x$ in the domain of $f$.
A function $f$ has a relative minimum at $c$ if $f(c) \leq f(x)$ for all $x$ in the vicinity of $c$. A function $f$ has a relative maximum at $c$ if $f(c) \geq f(x)$ for all $x$ in the vicinity of $c$.
Definition of a Critical Number
If $f^{\prime}(c)=0$ or if $\boldsymbol{f}^{\prime}(c)$ does not exist, then $c$ is called a critical number of $f$.
Test for Increasing and Decreasing Functions
1. A function $f$ is increasing on an interval if $f^{\prime}(x)>0$ on that interval.
2. A function $f$ is decreasing on an interval if $f^{\prime}(x)<0$ on that interval.
First Derivative Test
Let $c$ be a critical number of a function $f$ that is continuous on an open interval.
1. If $f^{\prime}(x)$ changes from negative to positive at $c$, then $f(c)$ is a relative minimum of $f$.
2. If $f^{\prime}(x)$ changes from positive to negative at $c$, then $f(c)$ is a relative maximum of $f$.
3. If $f^{\prime}(x)$ does not change signs at $c$, then $f(c)$ is neither a relative minimum nor a relative maximum.
Relative Extrema Occur Only at Critical Numbers
If $f$ has a relative minimum or a relative maximum at $x=c$, then $c$ is a critical number of $f$. But the converse of this theorem is not necessarily true. The critical numbers of a function need not produce relative extrema.
a.
The number 0 is a critical number because $f^{\prime}(0)$ does not exist, and the number 1 is a critical number because $f^{\prime}(1)=0$. The relative maximum is $f(0)=0$ and the relative minimum is $f(1)=-1$.
$f$ is increasing on $(-\infty, 0)$ and $(1, \infty)$.
$f$ is decreasing on $(0,1)$.
b.
The numbers $1 / 2$ and 2 are critical numbers because $f^{\prime}(1 / 2)=0$ and $f^{\prime}(2)=0$. The absolute minimum is $f(1 / 2)=-27 / 16$. Note that the critical number $x=2$ does not yield a relative minimum or a relative maximum. Not every critical number of a function produces a relative extrema. $f$ is decreasing on $(-\infty, 1 / 2)$ and $f$ is increasing on $(1 / 2, \infty)$.
c.
The number 0 is a critical number because $f^{\prime}(0)$ does not exist. The numbers -1 and 1 are critical numbers because $f^{\prime}(-1)=0$ and $f^{\prime}(1)=0$.
The relative maximum is $f(-1)=2$ and the relative minimum is $f(1)=-2$.
$f$ is increasing on $(-\infty,-1)$ and $(1, \infty)$.
$f$ is decreasing on $(-1,1)$.
Example
- Find the critical numbers of $f(x)=x^{1 / 3}(x-1)^2$.
▶️Answer/Explanation
Solution
$
\begin{aligned}
& f^{\prime}(x)=x^{1 / 3} \cdot 2(x-1)+\frac{1}{3} x^{-2 / 3}(x-1)^2=2 x^{1 / 3}(x-1)+\frac{1}{3 x^{2 / 3}}(x-1)^2 \\
& =\frac{6 x(x-1)+(x-1)^2}{3 x^{2 / 3}}=\frac{(7 x-1)(x-1)}{3 x^{2 / 3}} \\
& f^{\prime}(x) \text { does not exist when } x=0 . \\
& f^{\prime}(x)=0 \text { when } x=\frac{1}{7} \text { and } x=1 .
\end{aligned}
$
So, the critical numbers are $0, \frac{1}{7}$, and 1 .
The Extreme Value Theorem
If $f$ is continuous on the closed interval $[a, b]$, then $f$ has both a minimum and maximum on the interval.
Finding Extrema on a Closed Interval
To find the extrema of a continuous function $f$ on a closed interval $[a, b]$, use the following steps.
1. Find the values of $f$ at the critical numbers of $f$ in $(a, b)$.
2. Find the values of $f$ at the endpoints of the interval.
3. The least of these values is the minimum. The greatest is the maximum.
Example
- Find the absolute maximum and minimum values of $f(x)=x^4-2 x^3$ on the interval $[-1,3]$.
▶️Answer/Explanation
Solution
$\square \quad f^{\prime}(x)=4 x^3-6 x^2=2 x^2(2 x-3)$
Since $f^{\prime}(x)$ exists for all $x$, the only critical numbers of $f$ occurs when $x=0$ or $x=3 / 2$. $f(0)=0, f(3 / 2)=-1.6875$
The values of $f$ at the endpoints of the interval are $f(-1)=3$ and $f(3)=27$.
So, the absolute maximum value is $f(3)=27$ and the absolute minimum value is $f(3 / 2)=-1.6875$.
Example
- Let $f$ be the function given by $f(x)=x-2 \sin x$ for $0 \leq x \leq 2 \pi$.
(a) Find the intervals on which $f$ is increasing and decreasing.
(b) Find the absolute minimum and maximum value of $f$ on the closed interval $[0,2 \pi]$.
▶️Answer/Explanation
Solution
$
\begin{aligned}
& f^{\prime}(x)=1-2 \cos x=0 \\
& \Rightarrow \cos x=\frac{1}{2} \\
& \Rightarrow x=\frac{\pi}{3}, \frac{5 \pi}{3}
\end{aligned}
$
Since there are no points for which $f^{\prime}$ does not exists, we can conclude that $x=\frac{\pi}{3}$ and $x=\frac{5 \pi}{3}$ are the only critical numbers.
Make a diagram which shows the signs of $f^{\prime}$.
(a) $f$ is increasing on the interval $\left(\frac{\pi}{3}, \frac{5 \pi}{3}\right)$ and decreasing on the interval $\left(0, \frac{\pi}{3}\right)$ and $\left(\frac{5 \pi}{3}, 2 \pi\right)$.
(b) Absolute maximum value is 6.968 and absolute minimum value is -0.685 .
Exercises – The First Derivative Test and the Extreme Values of Functions
Multiple Choice Questions
- At what values of $x$ does $f(x)=(x-1)^3(3-x)$ have the absolute maximum?
(A) 1
(B) $\frac{3}{2}$
(C) 2
(D) $\frac{5}{2}$
▶️Answer/Explanation
Ans:D
Question
- At what values of $x$ does $f(x)=x-2 x^{2 / 3}$ have a relative minimum?
(A) $\frac{64}{27}$
(B) $\frac{16}{9}$
(C) $\frac{4}{3}$
(D) 2
▶️Answer/Explanation
Ans:A
Question
- What is the minimum value of $f(x)=x^2 \ln x$ ?
(A) $-e$
(B) $-\frac{1}{2 e}$
(C) $-\frac{1}{e}$
(D) $-\frac{1}{\sqrt{e}}$
▶️Answer/Explanation
Ans:B
Question
4. The graph of a function $f$ is shown above. Which of the following statements about $f$ are true?
I. $\lim _{x \rightarrow a} f(x)$ exists.
II. $x=a$ is the domain of $f$.
III. $f$ has a relative minimum at $x=a$.
(A) I only
(B) I and II only
(C) I and III only
(D) I, II, and III
▶️Answer/Explanation
Ans:D
Question
- A polynomial $f(x)$ has a relative minimum at $(-4,2)$, a relative maximum at $(-1,5)$, a relative minimum at $(3,-3)$ and no other critical points. How many zeros does $f(x)$ have?
(A) one
(B) two
(C) three
(D) four
▶️Answer/Explanation
Ans:B