Question

In Problems 21–32, find the derivative of each function at the given number.

$f\left(x\right)=x^3-2x^2+x$at$-1$

Short answer

The derivative of the given function at $-1$is equal to$8$

Step-by-step solution

Step 1. Given information

Given function is $f\left(x\right)=x^3-2x^2+x$

We have to find the derivative of the given function at$-1$

Step 2. Definition of the derivative

Let's say a function

The derivative of the function at is defined as

$f\text{'}\left(c\right)=\underset{x\to c}{\lim }\frac{f\left(x\right)-f\left(c\right)}{x-c}$

Step 3. Finding the derivative

By substituting the given function in the derivative definition,

we will get:

$$\begin{gathered} f\text{'}(-1)=\underset{x\to -1}{\lim }\frac{f\left(x\right)-f(-1)}{x-(-1)} \\ \Rightarrow f\text{'}(-1)=\underset{x\to -1}{\lim }\frac{(x^3-2x^2+x)-({(-1)}^3-2{(-1)}^2-1)}{x+1} \\ \Rightarrow f\text{'}(-1)=\underset{x\to -1}{\lim }\frac{(x^3-2x^2+x)+4}{x+1} \end{gathered}$$

Step 4. Simplifying the limit

By rewriting the limit, we will have:

$$\begin{gathered} f\text{'}(-1)=\underset{x\to -1}{\lim }\frac{x^3-2x^2+x+4}{x+1} \\ \Rightarrow f\text{'}(-1)=\underset{x\to -1}{\lim }\frac{(x+1)(x^2-3x+4)}{x+1} \\ \Rightarrow f\text{'}(-1)=\underset{x\to -1}{\lim }(x^2-3x+4) \\ \Rightarrow f\text{'}(-1)={(-1)}^2-3(-1)+4 \\ \Rightarrow f\text{'}(-1)=8 \end{gathered}$$