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Q21.

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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 .

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