Chapter 3.1, Problem 42E

(a)

To determine

Tocalculate:The value of $f\text{'}\left(x\right)$ for $x<1$ .

Answer to Problem 42E

The derivation is $f\text{'}\left(x\right)=2x$ for $x<1$ .

Explanation of Solution

Given information:

The given function: $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant

Calculation:

Since, $x<1$

Thus, the function will be $f\left(x\right)=x^2$

Find the derivatives of the function. Use the formula $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ .

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^2-x}{h}\\ =\underset{h\to 0}{\lim }\frac{x^2+2xh+h^2-x^2}{h}\\ =\underset{h\to 0}{\lim }\frac{2xh+h^2}{h}\\ =\underset{h\to 0}{\lim }\left[2x+h\right]\\ =2x\end{array}$

Hence, for $x<1$ the derivation is $f\text{'}\left(x\right)=2x$ .

(b)

To determine

To calculate:The value of $f\text{'}\left(x\right)$ for $x>1$ .

Answer to Problem 42E

The derivation is $f\text{'}\left(x\right)=2$ for $x>1$

Explanation of Solution

Given information:

The given function: $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant

Calculation:

It is given that $x>1$

Thus, the function will be $f\left(x\right)=2x$

Find the derivatives of the function. Use the formula $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ .

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{2\left(x+h\right)-2x}{h}\\ =\underset{h\to 0}{\lim }\frac{2x+2h-2x}{h}\\ =\underset{h\to 0}{\lim }\frac{2h}{h}\\ =\underset{h\to 0}{\lim }2\\ =2\end{array}$

Hence, for $x>1$ the derivation is $f\text{'}\left(x\right)=2$ .

(c)

To determine

To calculate:The value of $\underset{x\to 1^{-}}{\lim }f\text{'}\left(x\right)$ .

Answer to Problem 42E

The value of $\underset{x\to 1^{-}}{\lim }2x$ is 2.

Explanation of Solution

Given information:

The given function: $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant.

Calculation:

From 42a, for the value $x<1$ , the derivation $f\text{'}\left(x\right)=2x$ .

Thus substitute the value 1 for $x$ .

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{x\to 1^{-}}{\lim }2x\\ =2\times 1\\ =2\end{array}$

Hence, the value of $\underset{x\to 1^{-}}{\lim }2x$ is 2.

(d)

To determine

To calculate:The value of $\underset{x\to 1^{+}}{\lim }f\text{'}\left(x\right)$ .

Answer to Problem 42E

The value of $\underset{x\to 1^{+}}{\lim }2x$ is 2.

Explanation of Solution

Given information:

The given function: $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant.

Calculation:

From 42a, for the value $x>1$ , the derivation $f\text{'}\left(x\right)=2x$ .

Thus substitute the value 1 for $x$ .

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{x\to 1^{+}}{\lim }2x\\ =2\times 1\\ =2\end{array}$

Hence, the value of $\underset{x\to 1^{+}}{\lim }2x$ is 2.

(e)

To determine

Whether $\underset{x\to 1}{\lim }f\text{'}\left(x\right)$ exist or not.

Answer to Problem 42E

The limit $\underset{x\to 1}{\lim }f\text{'}\left(x\right)$ exist.

Explanation of Solution

Given information:

The given function: $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$

Formula Used:

Condition for existence of two-sided limit:

  $\underset{h\to 1^{-}}{\lim }f\text{'}\left(x\right)=\underset{h\to 1^{+}}{\lim }f\text{'}\left(x\right)$

Where $f\left(x\right)$ is the function.

Calculation:

It is known that, for two sided limit $\underset{h\to 1^{-}}{\lim }f\text{'}\left(x\right)=\underset{h\to 1^{+}}{\lim }f\text{'}\left(x\right)$ .

From 42c and 42d,

  $\underset{x\to 1^{-}}{\lim }2x=2$

And, $\underset{x\to 1^{+}}{\lim }2x=2$

Thus, $\underset{h\to 1^{-}}{\lim }f\text{'}\left(x\right)=\underset{h\to 1^{+}}{\lim }f\text{'}\left(x\right)=2$

Hence, the limit $\underset{x\to 1}{\lim }f\text{'}\left(x\right)$ exist.

(f)

To determine

To calculate:The left hand derivative of $f$ at $x=1$ .

Answer to Problem 42E

The limit $\underset{x\to 1}{\lim }f\text{'}\left(x\right)$ exist.

Explanation of Solution

Given information:

The given function: $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant.

Calculation:

The given function be $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$ .

Find the left hand derivative, use the formula $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ .

Write the above formula using the given function as:

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{h\to 0^{--}}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{{\left(1+h\right)}^2-1}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{1+2h+h^2-1}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{2h+h^2}{h}\\ =\underset{h\to 0^{-}}{\lim }\left[2+h\right]\\ =2\end{array}$

(g)

To determine

To calculate:The right hand derivative of $f$ at $x=1$ .

Answer to Problem 42E

The right hand derivative does not exist.

Explanation of Solution

Given information:

The given function: $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant.

Calculation:

The given function be $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$ .

For right sided limit, $f\left(1+h\right)$ is derived for $x>1$ . But for $f\left(1\right)$ use the derivation formula $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ .

Write the above formula using the given function as:

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{h\to 0^{+}}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{2\left(1+h\right)-1}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{2h+1}{h}\\ =\infty \end{array}$

Thus, the limit of $f\text{'}\left(x\right)$ for $x>1$ is undefined.

Hence, right hand derivative does not exist.

(h)

To determine

Whether the derivative $f\text{'}\left(1\right)$ exist or not.

Answer to Problem 42E

The right hand derivative does not exist.

Explanation of Solution

Given information:

The given function: $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant.

Calculation:

From 42f and 42g, the left hand derivative exist but the right hand derivative does not.

Hence, the derivative does not exist.