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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 $4(b+1)$?

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    $\begingroup$ Formatting tip: type $x_1$ to obtain $x_1$. $\endgroup$ – N. F. Taussig Aug 14 at 9:00
  • $\begingroup$ Thanks for the tip! $\endgroup$ – Doug Fir Aug 14 at 13:52
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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.

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There's a mistake.

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

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

Now it does work out.

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