Question

Find the different quotients of$f\left(x\right)=-2x^2+x+1$;that is,find $\frac{f(x+h)-f\left(h\right)}{h},h\ne 0$

Short answer

The difference quotient is$-4x+1-2h$

Step-by-step solution

Step 1.Given information

The given Function $f\left(x\right)=-2x^2+x+1$

we have to find $\frac{f(x+h)-f\left(x\right)}{h}$

Step 2.Compute the given function 

Substitute $(x+h)$for x in the equation,$f\left(x\right)=-2x^2+x+1$to obtain $f(x+h)$

$f(x+h)=-2{(x+h)}^2+(x+h)+1$

Now compute the function $\frac{f(x-h)-f\left(x\right)}{h}$

$$\begin{gathered} \frac{f(x+h)-f\left(x\right)}{h}=\frac{(-2{(x+h)}^2+(x+h)+1)-(-2x^2+x+1)}{h} \\ =\frac{(-2(x^2+2xh+h^2)+x+h+1)-(-2x^2+x+1)}{h} \\ =\frac{(-2x^2-4xh-2h^2+x+h+1+2x^2-x-1)}{h} \end{gathered}$$

Step 3.Combine the like terms and simplify

$$\begin{gathered} \frac{-2x^2-4xh-2h^2+x+h+1+2x^2-x-1}{h}=\frac{(2-2)x^2(1-1)x+(1-1)-4xh-2h^2+h}{h} \\ =\frac{-4xh+h+-2h^2}{h} \\ =-4xh+1-2h \end{gathered}$$

Therefore,the difference quotient is$-4x+1-2h$