Chapter 2.3, Problem 32AYU

In Problems 33-36, the graph of a function $f$ is given. Use the graph to find:

a. The numbers, if any, at which $f$ has a local maximum. What are the local maximum values?

b. The numbers, if any, at which $f$ has a local minimum. What are the local minimum values?

36.

To determine

To find: The following values using the given graph:

a. Local maximum points and its value.

b. Local minimum points and its value.

Answer to Problem 32AYU

From the graph, concluding the following results:

a. The curve has local maximum points at $x=0$. The value of the local maximum at $0$ is $f\left(0\right)=\text{ 1}$.

b. The curve has local minimum points at $x=-\pi$ and $x=\pi$. The value of the local minimum at $-\pi$ and $\pi$ are $f\left(-\pi \right)=-\text{1}$ and $f\left(\pi \right)=-\text{1}$.

Explanation of Solution

Given:

It is asked to find the local maximum and minimum points of the given function $f$ and its value.

Graph:

Interpretation:

a. Local maximum: By the definition of local maximum, Let $f$ be a function defined on some interval $I$.

A function $f$ has a local maximum at $c$ if there is an open interval $I$ containing $c$ so that, for all $x$ in this open interval, we have $f\left(x\right)\le f\left(c\right)$. We call $f\left(c\right)$ a local maximum value of $f$, It can be directly concluded from the graph and the definition that the curve has local maximum points at $x=0$.

The value of the local maximum at $0$ is $f\left(0\right)=\text{ 1}$.

b. Local minimum: By the definition of local minimum, Let $f$ be a function defined on some interval $I$.

A function $f$ has a local minimum at $c$ if there is an open interval $I$ containing $c$ so that, for all $x$ in this open interval, we have $f\left(x\right)\ge f\left(c\right)$. We call $f\left(c\right)$ a local minimum value of $f$, It can be directly concluded from the graph and the definition that the curve has local minimum points at $x=-\pi$ and $x=\pi$.

The value of the local minimum at $-\pi$ and $\pi$ are $f\left(-\pi \right)=-\text{1}$ and $f\left(\pi \right)=-\text{1}$.