# Rate of change with $f(x)=4x^2-7$ on $[1,b]$

by Doug Fir   Last Updated August 14, 2019 09:20 AM - source

As part of a textbook exercise I am to find the rate of change of $$f(x)=4x^2-7$$ on inputs $$[1,b]$$.

The solution provided is $$4(b+1)$$ and I am unable to arrive at this solution.

Tried:

$$f(x_2)=4b^2-7$$

$$f(x_1)=4(1^2)-7=4-7=-3$$

If the rate of change is $$\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ then: $$\frac{(4b^2-7)-3}{b-1}$$ = $$\frac{4b^2-10}{b-1}$$

This is as far as I got. I tried to see if I could factor out the numerator but this didn't really help me:

$$(4b^2-10)=2(2b^2-5)$$

If I substitute this for my numerator I still cannot arrive at the provided solution.

I then tried isolating b in the numerator: $$4b^2-10=0$$

$$4b^2=10$$

$$b^2=10/4$$

$$b=\frac{\sqrt{10}}{\sqrt{4}}=\frac{\sqrt{10}}{2}$$

This still doesn't help me arrive at the solution.

How can I arrive at $$4(b+1)$$?

Tags :

There's a mistake.

$$f(x_2) - f(x_1) = 4b^2-7 - (-3) =4b^2 -4 = 4(b+1)(b-1)$$

A small sign-mistake! See the highlighted parts in red and blue:

Tried:

$$f(x_2)=4b^2-7$$

$$\color{blue}{f(x_1)}=4(1^2)-7=4-7=\color{blue}{-3}$$

If the rate of change is $$\frac{f(x_2)-f(x_1)}{x_2-x_1}$$ then: $$\frac{(4b^2-7)\color{red}{-3}}{b-1}$$ = $$\frac{4b^2-10}{b-1}$$

Which should be:

$$\frac{f(x_2)\color{red}{-}\color{blue}{f(x_1)}}{x_2-x_1} = \frac{(4b^2-7)\color{red}{-}\left(\color{blue}{-3}\right)}{b-1} = \frac{4b^2-4}{b-1}$$

Then proceed with $$4b^2-4=4\left(b^2-1\right)=4\left(b-1\right)\left(b+1\right)$$ and simplify.

Your $$f(1)$$ is $$-3$$ so $$f(b)-f(1)=4b^2-7+3=4b^2-4$$

Now it does work out.