Chapter 2.5, Problem 37E

To determine

To calculate: The value of $b=1$

Answer to Problem 37E

The value of $b$ cannot be $1$ when the function is both continuous and differentiable at $x=0$ .

Explanation of Solution

Given information:

The function is. $g\left(x\right)=\left\{\begin{array}{c}x+b,x<0\\ \cos x,x\ge 0\end{array}\right\}$

Concept used:

The function will be continuous at a point if the right hand limit, left hand limit and limit at that point are equal.

The function is said to be differentiable when the right hand derivative and left hand derivative are equal.

Calculation:

The function is $g\left(x\right)=\left\{\begin{array}{c}x+b,x<0\\ \cos x,x\ge 0\end{array}\right\}$ .

Continuity at $x=0$

The limit at $x=0$is as shown below.

  $\begin{array}{c}g\left(0\right)=\cos \left(0\right)\\ =1\end{array}$

The right hand limit is as shown below.

  $\begin{array}{c}\underset{x\to 0^{+}}{\lim }g\left(x\right)=\cos \left(0\right)\\ =1\end{array}$

The left hand limit is as shown below.

  $\begin{array}{c}\underset{x\to 0^{-}}{\lim }g\left(x\right)=x+b\\ =0+b\\ =b\end{array}$

Equate the limits for the value of $b$ .

  $b=1$

Substitute $1$ for $b$ then the function becomes $g\left(x\right)=\left\{\begin{array}{c}x+1,x<0\\ \cos x,x\ge 0\end{array}\right\}$

Check the differentiability for the function $g\left(x\right)=\left\{\begin{array}{c}x+1,x<0\\ \cos x,x\ge 0\end{array}\right\}$ at $x=0$ .

The right hand derivative is as shown below.

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{g\left(a+h\right)-g\left(a\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{\cos \left(0+h\right)-1}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{\cosh -1}{h}\\ =0\end{array}$

The left hand derivative is as shown below.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{g\left(a+h\right)-g\left(a\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{\left(0+1+h\right)-1}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{1+h-1}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{h}{h}\\ =1\end{array}$

Conclusion: The function is not differentiable at $x=0$ when $b=1$ .