Chapter 4, Problem 57RE

a.

To determine

Determine where the function is differentiable.

Answer to Problem 57RE

Differentiable for $(-1,1)\cup (1,3)$ .

Explanation of Solution

Given information: Given function is

  $\begin{array}{l}f(x)=\sqrt{x^2+3},\text{if}-1\le x<1.\\ f(x)=x+1,\text{if}1\le x<3\end{array}$

Calculation:

  $f(x)=\sqrt{x^2+3}$ is continuous for real values of xsince $x^2+3\ge 0$ for all x . $f(x)=x+1$ is continuous for all x since linear function are continuous for all real numbers. The only possible location of a discontinuity is then x =1 since $f(x)$ switches from $\sqrt{x^2+3}tox+1$ at this value. To determine if $f(x)$ is continuous at x =1, we must show $f(1)\text{exists}\underset{x\to 1}{\lim }f(x)\text{exists}\text{and}\underset{x\to 1}{\lim }f(x)=f(1).$

When $\begin{array}{l}x=1,f(1)=1+1=2.\\ f(1)=2.\end{array}$

  $\begin{array}{l}\underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{-}}{\lim }\sqrt{x^2+3}=\sqrt{1^2+3}=\sqrt{4}=2.\\ \underset{x\to 1^{+}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }(x+1)=1+1=2\end{array}$

The left and right hand limits both equal 2 so $\underset{x\to 1}{\lim }f(x)=2$.Therefore $\underset{x\to 1}{\lim }f(x)=2=f(1).$

So, $f(x)$ is continuous at x =1.

The function is then continuous for (-1,3) since the two pieces of the domain were (-1,1) and (1,3) and it was continuous at x =1.

b.

To determine

Determine where the function is continuous but not differentiable.

Answer to Problem 57RE

Continuous but not differentiable at x =1. .

Explanation of Solution

Given information: Given function is

  $\begin{array}{l}f(x)=\sqrt{x^2+3},\text{if}-1\le x<1.\\ f(x)=x+1,\text{if}1\le x<3\end{array}$

Calculation:

  $f(x)$ is continuous on its entire domain of (-1,3) so to determine if there are any values of x where $f(x)$ is not differentiable we must determine if there are any values of x that makes $f\text{'}(x)$ discontinuous .

  $\begin{array}{l}f(x)=\sqrt{x^2+3},\text{if}-1\le x<1.\\ f(x)=x+1,\text{if}1\le x<3.\\ $ f\text{'}(x)=\frac{1}{2}{(x^2+3)}^{-\frac{1}{2}}.2x,\text{if}-1\le x<1.$f\text{'}(x)=1,\text{if}1<x<3.\\ f\text{'}(x)=\frac{x}{\sqrt{x^2+3}},\text{if}-1\le x<1.\\ f\text{'}(x)=1,\text{if}1<x<3.\\ \underset{x\to 1^{-}}{\lim }f\text{'}(x)=\underset{x\to 1^{-}}{\lim }\frac{x}{\sqrt{x^2+3}}.\\ \underset{x\to 1^{-}}{\lim }f\text{'}(1)=\underset{x\to 1^{-}}{\lim }\frac{1}{\sqrt{1^2+3}}=\frac{1}{2}\\ \underset{x\to 1+}{\lim }f\text{'}(x)=\underset{x\to 1^{+}}{\lim }1\\ \underset{x\to 1+}{\lim }f\text{'}(1)=\underset{x\to 1^{+}}{\lim }1=1.\\ \end{array}$

The left and right hand limits are not equal so $f\text{'}(x)$ is discontinuous at x =1.Therfore $f(x)$ is not differentiable at x =1.

c.

To determine

Determine where the function neither continuous nor differentiable.

Answer to Problem 57RE

Neither continuous nor differentiable,nowhere.

Explanation of Solution

Given information: Given function is

  $\begin{array}{l}f(x)=\sqrt{x^2+3},\text{if}-1\le x<1.\\ f(x)=x+1,\text{if}1\le x<3\end{array}$

Calculation:

In summary $f(x)$ neither continuous nor differentiable no where.