Question

Let f(X) be defines for all x by the equation f(x)= 12x+8. Thus, f(2)=32. If f(x) / f(0)=2x, then find the?
value of x. I'm so confused with this problem. Someone please help me solve this problem!


Please show work and explain each step because I want to understand how to the problem

Answer 1

what is the question asking for?

Answer 2

Okay, first of all, we have a defined function
\[\Large f(\color{red}x)=12\color{red}x + 8 \] right?

Answer 3

yes

Answer 4

the answer is supposed to be 2

Answer 5

Let's find out.
Your problem also tells you (though only for the sake of information) that
\[\Large f(\color{red}2) = 12(\color{red}2)+ 8 = 24+8 = \color{blue}{32}\]

Answer 6

Now, what is the value of \(\large f(\color{red}0) = \color{green}?\)

Answer 7

that's what confused me. How do i find that out?

Answer 8

Well, when you have \[\Large f(\color{red}x) = <some \ expression \ involving \ \color{red}x>\]

Then, say \[\Large f(\color{red}a) \] just asks you to replace all the x's with a's.

Just like in this example, we have
\[\Large f(\color{red}x) = 12\color{red}x+8\]
\[\Large f(\color{red}2) = 12(\color{red}2) + 8=\color{red}{24}+8 = \color{blue}{32}\]

Answer 9

As you can see, to get the value of \(\large f(\color{red}2)\), we simply replaced the \(\large \color{red}x\) in \(\large 12\color{red}x+8\) with a \(\large \color{red}2\) and then solved :)

Answer 10

so for f(0) it would be 12(0) +8=8

Answer 11

YES. Exactly :)

Answer 12

So, we are given..
\[\Large \frac{f(\color{red}x)}{f(\color{red}0)}=2\color{red}x\]

Answer 13

\(\large f(\color{red}x)\) is just \(\large 12\color{red}x+8\) so we replace it with that...

\[\Large \frac{12\color{red}x+8}{f(\color{red}0)}= 2\color{red}x\]

Answer 14

What's \(\large f(\color{red}0)\) again? ;)

Answer 15

8

Answer 16

That's right :)
So we replace it with 8.
\[\Large \frac{12\color{red}x+8}{8 }= 2\color{red}x\]

Could you solve for x here? :)

Answer 17

12x +8 = 16x 8=4x x=2

Answer 18

Good job :)

Answer 19

Thank you so much ! :))

Answer 20

No problem :)