Chapter 4.1, Problem 12QR

(a)

To determine

To find : domain of $f^{\prime }$ .

Answer to Problem 12QR

The domain of the function $f^{\prime }$ is $\left(-\infty ,\infty \right)$ .

Explanation of Solution

Given information :

  $f\left(x\right)=\left\{\begin{array}{l}{x}^3-2x,x\le 2\\ x+2,x>2\end{array}\right.$

Differentiate $f\left(x\right)=x^3-2x$ with respect to $x$ ,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(x^3-2x\right)\\ =\frac{d}{dx}\left(x^3\right)-\frac{d}{dx}\left(2x\right)\\ =\left(3x^2\right)-\left(2\right)\\ =3x^2-2\end{array}$

Therefore, $f^{\prime }\left(x\right)=3x^2-2,x\le 2$ .

Differentiate $f\left(x\right)=x+2$ with respect to $x$ ,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(x+2\right)\\ =\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(2\right)\\ =1\end{array}$

Therefore, $f^{\prime }\left(x\right)=1,x>2$ .

Hence, the domain of the function $f^{\prime }$ is $\left(-\infty ,\infty \right)$ .

(b)

To determine

To find : formula for $f^{\prime }\left(x\right)$ .

Answer to Problem 12QR

  $f^{\prime }\left(x\right)=\left\{\begin{array}{l}3x^2-2,x\le 2\\ 1,x>2\end{array}\right.$ .

Explanation of Solution

Given information :

  $f\left(x\right)=\left\{\begin{array}{l}{x}^3-2x,x\le 2\\ x+2,x>2\end{array}\right.$

Differentiate $f\left(x\right)=x^3-2x$ with respect to $x$ ,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(x^3-2x\right)\\ =\frac{d}{dx}\left(x^3\right)-\frac{d}{dx}\left(2x\right)\\ =\left(3x^2\right)-\left(2\right)\\ =3x^2-2\end{array}$

Therefore, $f^{\prime }\left(x\right)=3x^2-2,x\le 2$ .

Differentiate $f\left(x\right)=x+2$ with respect to $x$ ,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(x+2\right)\\ =\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(2\right)\\ =1\end{array}$

Therefore, $f^{\prime }\left(x\right)=1,x>2$ .

Hence, $f^{\prime }\left(x\right)=\left\{\begin{array}{l}3x^2-2,x\le 2\\ 1,x>2\end{array}\right.$ .