Suggested languages for you:

Americas

Europe

Q21.

Expert-verified
Found in: Page 922

Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

Find the derivative of the function at the given number.$f\left(x\right)=x-3{x}^{2},\text{}at\text{}-1$.

The derivative of the function at $-1$is7.

See the step by step solution

Step 1. Given information.

The function here given is,

$f\left(x\right)=x-3{x}^{2},\text{}at\text{}-1$

Step 2. Formula used.

The derivative of a function f at a number, denoted by$f\text{'}\left(a\right),$ is

.$f\text{'}\left(a\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(a+h\right)-f\left(a\right)}{h}$

if this limit exists.

Step 3. Finding the slope of the tangent line at a given point.

According to the definition of a derivative, with,$a=-1$ we have

$\begin{array}{c}f\text{'}\left(-1\right)=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(-1+h\right)-f\left(-1\right)}{h}\text{}Definition\text{}of\text{}f\text{'}\left(-1\right)\\ =\underset{h\to 0}{\mathrm{lim}}\frac{\left[\left(-1+h\right)-3{\left(-1+h\right)}^{2}\right]-\left[-1-3{\left(-1\right)}^{2}\right]}{h}\text{}f\left(x\right)=x-3{x}^{2}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{\left[-1+h-3\left(1+{h}^{2}-2h\right)\right]+4}{h}\text{}Expand\\ =\underset{h\to 0}{\mathrm{lim}}\frac{-1+h-3-3{h}^{2}+6h+4}{h}\\ =\underset{h\to 0}{\mathrm{lim}}\frac{-3{h}^{2}+7h}{h}\text{}Simplify\\ =\underset{h\to 0}{\mathrm{lim}}\left(-3h+7\right)\text{}Cancel\text{}h\\ =7\end{array}$

Therefore, the derivative of the function at $-1$is7 .