Chapter 13, Problem 4T

(a)

To determine

To find: The value of $f\text{'}\left(x\right)$ when value of function $f\left(x\right)$ is $x^2-2x$.

Answer to Problem 4T

The value of $f\text{'}\left(x\right)$ is $2x-2$.

Explanation of Solution

Given:

The value of function $f\left(x\right)$ is,

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

Calculation:

The derivative of the function $f\left(x\right)$ is,

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ (1)

The value of function $f\left(x\right)$ is,

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

Substitute $\left(x+h\right)$ for x in the above function.

$f\left(x+h\right)={\left(x+h\right)}^2-2\left(x+h\right)$

Substitute the values of $f\left(x+h\right)={\left(x+h\right)}^2-2\left(x+h\right)$ and $f\left(x\right)=x^2-2x$ in equation (1) to find the value of $f\text{'}\left(x\right)$,

$\begin{array}{c}f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\left(\frac{{\left(x+h\right)}^2-2\left(x+h\right)-x^2+2x}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{x^2+h^2+2xh-2x-2h-x^2+2x}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h^2+2xh-2h}{h}\right)\\ =\underset{h\to 0}{\lim }\left(h+2x-2\right)\end{array}$

Apply limit of sum and difference,

$\begin{array}{c}f\text{'}\left(x\right)=\left(0+2x-2\right)\\ =2x-2\end{array}$

Thus the value of $f\text{'}\left(x\right)$ is $2x-2$.

(b)

To determine

To find: The values of $f\text{'}\left(-1\right)$, $f\text{'}\left(1\right)$ and $f\text{'}\left(2\right)$ when value of function $f\left(x\right)$ is $x^2-2x$.

Answer to Problem 4T

The values of $f\text{'}\left(-1\right)$ is $\underset{\_}{-4}$, $f\text{'}\left(1\right)$ is 0 and $f\text{'}\left(2\right)$ is 2.

Explanation of Solution

Given:

The value of function $f\left(x\right)$ is,

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

Calculation:

From part (a), the value of $f\text{'}\left(x\right)$ is $2x-2$.

Substitute $-1$ for x in derivative function $f\text{'}\left(x\right)$ for $f\text{'}\left(-1\right)$,

$\begin{array}{c}f\text{'}\left(-1\right)=2\cdot \left(-1\right)-2\\ =-2-2\\ =-4\end{array}$

Substitute $1$ for x in derivative function $f\text{'}\left(x\right)$ for $f\text{'}\left(1\right)$,

$\begin{array}{c}f\text{'}\left(1\right)=2\cdot \left(1\right)-2\\ =2-2\\ =0\end{array}$

Substitute 2 for x in derivative function $f\text{'}\left(x\right)$ for $f\text{'}\left(2\right)$,

$\begin{array}{c}f\text{'}\left(2\right)=2\cdot \left(2\right)-2\\ =4-2\\ =2\end{array}$

Thus, the values of $f\text{'}\left(-1\right)$ is $\underset{\_}{-4}$, $f\text{'}\left(1\right)$ is 0 and $f\text{'}\left(2\right)$ is 2.