Q. 78 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 75– 82, find the difference quotient of $f$; that is, find $\frac{f(x+h)- f\left(x\right)}{h}, h\ne 0,$for each function. Be sure to simplify.

$f\left(x\right)=3x^2-2x+6$

Step-by-Step Solution

Answer

The value of $\frac{f(x+h)- f\left(x\right)}{h}=6x+3h-2$

Step 1. Given Information

In this problems we have to find the difference quotient of $f$; that is, find $\frac{f(x+h)- f\left(x\right)}{h}, h\ne 0,$for given function. Be sure to simplify.

The given function is
$f\left(x\right)=3x^2-2x+6$

Step 2. Firstly we find the value of f ( x + h ) We find the value of f ( x + h ) by putting x = x + h in given function.

$f(x+h)=3{(x+h)}^2-2(x+h)+6$

Using the formula ${(a+b)}^2=a^2+2ab+b^2$

$$\begin{gathered} f(x+h)=3\left({x^2+2xh+h}^2\right)-2(x+h)+6 \\ f(x+h)=3{x^2+6xh+3h}^2-2x-2h+6 \end{gathered}$$

Step 3. Putting the value of f ( x )   and   f ( x + h ) in the f ( x + h ) -   f ( x ) h

$$\begin{gathered} \frac{f(x+h)- f\left(x\right)}{h}=\frac{3{x^2+6xh+3h}^2-2x-2h+6-(3x^2-2x+6)}{h} \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{3{x^2+6xh+3h}^2-2x-2h+6-3x^2+2x-6}{h} \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{{6xh+3h}^2-2h}{h} \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{h(6x+3h-2)}{h} \\ \frac{f(x+h)- f\left(x\right)}{h}=6x+3h-2 \end{gathered}$$