Chapter 4, Problem 56RE

a.

To determine

Determine where the function is differentiable.

Answer to Problem 56RE

For all real values of x .

Explanation of Solution

Given information: Given function is $g(x)=\sin (x^2+1).$

Calculation:

First determine if y is continuous everywhere. $x^2+1$ is continuous for all values of x since it’s a polynomial . $\sin x$ is also continuous for all values of x .The composition of $\sin (x^2+1)$ must also then be continuous for all real values of x .

b.

To determine

Determine where the function is continuous but not differentiable.

Answer to Problem 56RE

Nowhere.

Explanation of Solution

Given information: Given function is $g(x)=\sin (x^2+1).$

Calculation:

Next determine if y will be differentiable everywhere if $y\text{'}$ is continuous everywhere .The derivative of ,

  $\begin{array}{l}g(x)=y=\sin (x^2+1).\\ y\text{'}=2x\cos (x^2+1).\end{array}$

2x is continuous for all values of x since its linear and $\cos xand{x}^2+1$ are both continuous for all real numbers so $\cos (x^2+1)$ is also continuous for all real numbers. The product $2x\cos (x^2+1)$ must then also be continuous for all real numbers .Since $y\text{'}$ is then continuous for all real numbers , y is differentiable for all real numbers.

c.

To determine

Determine where the function neither continuous nor differentiable.

Answer to Problem 56RE

Nowhere.

Explanation of Solution

Given information: Given function is $g(x)=\sin (x^2+1).$

Calculation:

Next determine if y will be differentiable everywhere if $y\text{'}$ is continuous everywhere .The derivative of ,

  $\begin{array}{l}g(x)=y=\sin (x^2+1).\\ y\text{'}=2x\cos (x^2+1).\end{array}$

2x is continuous for all values of x since its linear and $\cos xand{x}^2+1$ are both continuous for all real numbers so $\cos (x^2+1)$ is also continuous for all real numbers. The product $2x\cos (x^2+1)$ must then also be continuous for all real numbers .Since $y\text{'}$ is then continuous for all real numbers , y is differentiable for all real numbers.