Chapter 4.1, Problem 54E

(a.)

To determine

To Show: There is an open interval containing $c$ such that $f\left(x\right)-f\left(c\right)\le 0$ for all $x$ in the open interval.

Answer to Problem 54E

It has been shown that there is an open interval containing $c$ such that $f\left(x\right)-f\left(c\right)\le 0$ for all $x$ in the open interval.

Explanation of Solution

Given:

The function $f$ has a local maximum value at the interior point $c$ of its domain and $f^{\prime }\left(c\right)$ exists.

Concept used:

If a function $f$ has a local maximum value at the interior point $c$ of its domain, then $f\left(c\right)\ge f\left(x\right)$ for all $x\in \left(c-a,c+a\right)$ for some $a>0$ .

Calculation:

It is given that the function $f$ has a local maximum value at the interior point $c$ of its domain.

As discussed, there exists some $a>0$ such that $f\left(c\right)\ge f\left(x\right)$ for all $x\in \left(c-a,c+a\right)$ .

Simplifying, $f\left(x\right)-f\left(c\right)\le 0$ for all $x\in \left(c-a,c+a\right)$ .

This shows that there is an open interval containing $c$ such that $f\left(x\right)-f\left(c\right)\le 0$ for all $x$ in the open interval.

Conclusion:

It has been shown that there is an open interval containing $c$ such that $f\left(x\right)-f\left(c\right)\le 0$ for all $x$ in the open interval.

(b.)

To determine

To Explain: Why $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$ .

Answer to Problem 54E

It has been explained why $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$ .

Explanation of Solution

Given:

The function $f$ has a local maximum value at the interior point $c$ of its domain and $f^{\prime }\left(c\right)$ exists.

Concept used:

If $x\to c^{+}$ , then $x>c$ .

Calculation:

As shown previously, there is an open interval containing $c$ such that $f\left(x\right)-f\left(c\right)\le 0$ for all $x$ in the open interval.

Then, $f\left(x\right)-f\left(c\right)\le 0$ for $x\to c^{+}$ .

Now, if $x\to c^{+}$ , then $x>c$ .

Then, $x-c>0$ for $x\to c^{+}$ .

Combining, it follows that $\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$ for $x\to c^{+}$ .

This shows that $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$ .

Conclusion:

It has been explained why $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$ .

(c.)

To determine

To Explain: Why $\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$ .

Answer to Problem 54E

It has been explained why $\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$ .

Explanation of Solution

Given:

The function $f$ has a local maximum value at the interior point $c$ of its domain and $f^{\prime }\left(c\right)$ exists.

Concept used:

If $x\to c^{-}$ , then $x<c$ .

Calculation:

As shown previously, there is an open interval containing $c$ such that $f\left(x\right)-f\left(c\right)\le 0$ for all $x$ in the open interval.

Then, $f\left(x\right)-f\left(c\right)\le 0$ for $x\to c^{-}$ .

Now, if $x\to c^{-}$ , then $x<c$ .

Then, $x-c<0$ for $x\to c^{-}$ .

Combining, it follows that $\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$ for $x\to c^{-}$ .

This shows that $\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$ .

Conclusion:

It has been explained why $\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$ .

(d.)

To determine

How parts (b) and (c) help to conclude that $f^{\prime }\left(c\right)=0$ .

Answer to Problem 54E

It has been explained how parts (b) and (c) help to conclude that $f^{\prime }\left(c\right)=0$ .

Explanation of Solution

Given:

The function $f$ has a local maximum value at the interior point $c$ of its domain and $f^{\prime }\left(c\right)$ exists.

Concept used:

According to definition of derivative,

  $f^{\prime }\left(c\right)=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$

According to the property of limit,

  $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$

Calculation:

It is given that $f^{\prime }\left(c\right)$ exists.

Since $f^{\prime }\left(c\right)=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ , it follows that $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ exists.

Since $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ , it follows that,

  $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ .

As shown previously, $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$ and $\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$ .

Then, $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ implies that,

  $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=0$

Put $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=0$ in $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ to get,

  $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=0$

Put $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=0$ in $f^{\prime }\left(c\right)=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ to get,

  $f^{\prime }\left(c\right)=0$

This explains how parts (b) and (c) help to conclude that $f^{\prime }\left(c\right)=0$ .

Conclusion:

It has been explained how parts (b) and (c) help to conclude that $f^{\prime }\left(c\right)=0$ .

(e.)

To determine

A similar argument if $f$ has a local minimum value at an interior point.

Answer to Problem 54E

A similar argument for $f^{\prime }\left(c\right)=0$ if $f$ has a local minimum value at an interior point has been provided.

Explanation of Solution

Given:

The function $f$ has a local maximum value at the interior point $c$ of its domain and $f^{\prime }\left(c\right)$ exists.

Concept used:

If a function $f$ has a local minimum value at the interior point $c$ of its domain, then $f\left(c\right)\le f\left(x\right)$ for all $x\in \left(c-a,c+a\right)$ for some $a>0$ .

Calculation:

It is given that the function $f$ has a local minimum value at the interior point $c$ (say) of its domain.

As discussed, there exists some $a>0$ such that $f\left(c\right)\le f\left(x\right)$ for all $x\in \left(c-a,c+a\right)$ .

Simplifying, $f\left(x\right)-f\left(c\right)\ge 0$ for all $x\in \left(c-a,c+a\right)$ .

Then, $f\left(x\right)-f\left(c\right)\ge 0$ for $x\to c^{+}$ .

Now, if $x\to c^{+}$ , then $x>c$ .

Then, $x-c>0$ for $x\to c^{+}$ .

Combining, it follows that $\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$ for $x\to c^{+}$ .

So, $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$ .

As shown previously, $f\left(x\right)-f\left(c\right)\ge 0$ for all $x\in \left(c-a,c+a\right)$ .

Then, $f\left(x\right)-f\left(c\right)\ge 0$ for $x\to c^{-}$ .

Now, if $x\to c^{-}$ , then $x<c$ .

Then, $x-c<0$ for $x\to c^{-}$ .

Combining, it follows that $\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$ for $x\to c^{-}$ .

So, $\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$ .

Now, it is given that $f^{\prime }\left(c\right)$ exists.

Since $f^{\prime }\left(c\right)=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ , it follows that $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ exists.

Since $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ , it follows that,

  $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ .

As shown previously, $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\ge 0$ and $\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}\le 0$ .

Then, $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ implies that,

  $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=0$

Put $\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=0$ in $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{+}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{-}}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ to get,

  $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=0$

Put $\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}=0$ in $f^{\prime }\left(c\right)=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$ to get,

  $f^{\prime }\left(c\right)=0$

Conclusion:

A similar argument for $f^{\prime }\left(c\right)=0$ if $f$ has a local minimum value at an interior point has been provided.