Chapter 14, Problem 33RE

To determine

To find: The value of function $f\left(x\right)=-4x^2+5$ at $0$ .

Answer to Problem 33RE

The derivative of $f\left(x\right)=-4x^2+5$ when $c=3$ is $f^{\text{'}}\left(3\right)=-24$ .

Explanation of Solution

Given information:

The given function is $f\left(x\right)=-4x^2+5$ at $0$ .

Calculation:

Calculate the function,

  $f\left(x\right)=-4x^2+5$ , use those to find $f\left(c\right)$ .

Substitute the value of $c=3$ .

  $\begin{array}{c}f\left(x\right)=-4x^2+5\\ =-4{\left(3\right)}^2+5\\ =-36+5\\ =-31\end{array}$

Substitute the value of $c=3$ , $f\left(c\right)=-31$ in function $f\left(x\right)=-4x^2+5$ .

  $\begin{array}{c}{f}^{\text{'}}\left(1\right)=\underset{x\to 3}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-\left(3\right)}\\ =\underset{x\to 3}{\lim }\frac{-4x^2+5-\left(-31\right)}{x-3}\\ =\underset{x\to 3}{\lim }\frac{-4\left(x-3\right)\left(x+3\right)}{x}\\ =\underset{x\to 3}{\lim }\left(-4x-12\right)\end{array}$

  $\begin{array}{c}{f}^{\text{'}}\left(1\right)=\left(-4\left(3\right)-12\right)\\ =-12-12\\ =-24\end{array}$

Therefore the derivative of $f\left(x\right)=-4x^2+5$ when $c=3$ is $f^{\text{'}}\left(3\right)=-24$ .