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 $4(b+1)$?
$\endgroup$ 23 Answers
$\begingroup$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.
$\endgroup$ 0 $\begingroup$There's a mistake.
$$f(x_2) - f(x_1) = 4b^2-7 - (-3) =4b^2 -4 = 4(b+1)(b-1)$$
$\endgroup$ $\begingroup$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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