Question

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

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

Short answer

Derivative of the function at$2$is equal to$20$

Step-by-step solution

Step 1. Given information

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

We have to find the derivative of the given function at$2$

Step 2. Definition of derivative

Let's say a function $y=f\left(x\right)$,

The derivative of the function at $c$is defined as

localid="1647080078052">f'(c)=limx→cf(x)-f(c)x-c.

Step 3. Finding the derivative

By substituting the given function in the derivative definition,

we will get:

$$\begin{gathered} f\text{'}\left(2\right)=\underset{x\to 2}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2} \\ \Rightarrow f\text{'}\left(2\right)=\underset{x\to 2}{\lim }\frac{(2x^3-x^2)-(2·2^3-2^2)}{x-2} \\ \Rightarrow f\text{'}\left(2\right)=\underset{x\to 2}{\lim }\frac{(2x^3-x^2)-12}{x-2} \end{gathered}$$

Step 4. Simplifying the limit

By rewriting the limit we will have:

$$\begin{gathered} f\text{'}\left(2\right)=\underset{x\to 2}{\lim }\frac{2x^3-x^2-12}{x-2} \\ \Rightarrow f\text{'}\left(2\right)=\underset{x\to 2}{\lim }\frac{(x-2)(2x^2+3x+6)}{x-2} \\ \Rightarrow f\text{'}\left(2\right)=\underset{x\to 2}{\lim }(2x^2+3x+6) \\ \Rightarrow f\text{'}\left(2\right)=2\left(4\right)+3\left(2\right)+6 \\ \Rightarrow f\text{'}\left(2\right)=20 \end{gathered}$$