Question

In Problems $33–35$, find the derivative of each function at the number indicated.

$\text{33.}f\left(x\right)=-4x^2+5$at$3$

Short answer

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

Step-by-step solution

Step 1. Find f(3)

To find the derivative we have to use the formula,

$f^{\mathrm{\prime }}\left(c\right)=\underset{x\to c}{lim}\frac{f\left(x\right)-f\left(c\right)}{x-c}$

We have been given $f\left(x\right)=-4x^2+5$and $c=3$

Now we have to find$f\left(c\right)$, therefore substitute $c=3$in $f\left(x\right)$to find$f\left(3\right)$.

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

Step 2. Use the derivative formula

Substituting all the values in the derivative formula we get,

$$\begin{gathered} \begin{array}{rl}{f}^{\mathrm{\prime }}\left(3\right)& =\underset{x\to 3}{lim}\frac{f\left(x\right)-f\left(3\right)}{x-\left(3\right)}\\ & =\underset{x\to 3}{lim}\frac{-4x^2+5-(-31)}{x-3}\\ & =\underset{x\to 3}{lim}\frac{-4x^2+36}{x-3}\\ & =\underset{x\to 3}{lim}\frac{-4(x-3)(x+3)}{x-3}\end{array} \\ \begin{array}{rl}& =\underset{x\to 3}{lim}(-4x-12)\\ & =(-4(3)-12)\\ & =-12-12\\ & =-24\end{array} \end{gathered}$$

Hence the derivative of $f\left(x\right)$when$c=3$is$f^{\mathrm{\prime }}\left(3\right)=-24$.