Chapter 2, Problem 11CT

Find the difference quotient of $f(x)=x^2-3x.$

To determine

To calculate: The difference quotient of the function $f(x)=x^2-3x$.

Answer to Problem 11CT

Solution:

The difference quotient of the function $f(x)=x^2-3x$ is $2x+h-3$.

Explanation of Solution

Given Information:

The function is $f(x)=x^2-3x$.

Explanation:

The difference quotient of function is $\frac{f(x+h)-f(x)}{h}$.

To find $f(x+h)$, replace $x$ by $x+h$ in $f(x)=x^2-3x$,

$\Rightarrow f(x+h)={(x+h)}^2-3(x+h)$,

$\Rightarrow f(x+h)=x^2+2xh+h^2-3x-3h$.

Substitute the value of $f(x+h)$ and $f(x)$ in $\frac{f(x+h)-f(x)}{h}$,

$\Rightarrow \frac{f(x+h)-f(x)}{h}=\frac{x^2+2xh+h^2-3x-3h-x^2+3x}{h}$,

Combine like terms,

$\Rightarrow \frac{f(x+h)-f(x)}{h}=\frac{2xh+h^2-3h}{h}$.

Factor out $h$ from above equation,

$\Rightarrow \frac{f(x+h)-f(x)}{h}=\frac{h(2x+h-3)}{h}$.

Cancel $h$ from above equation,

$\Rightarrow \frac{f(x+h)-f(x)}{h}=2x+h-3$.

Therefore, the difference quotient is $2x+h-3$.