Question

I got stuck halfway through a problem, here's what I have so far:

Suppose that f(x) is a function with f(2x+4) = 6x+3. Find f(5-3x) + f(5x-1).

I'm thinking that I plug in (2/5)x + 1 into f(5x-1) to get 6x +3 but I was unsure of whether or not to plug that in again.

Thanks so much!

Answer 1

first step:
\[f(2x+4)=6x+3 \\ u=2x+4 \\ \text{ solve for } x \\ \text{ write } x \text{ \in terms of } u \text{ given } u=2x+4\]

Answer 2

X in terms of u is (-u+4)/2 but I'm not sure what to do with that ;A;

Answer 3

\[x=\frac{u-4}{2} \\ \text{ given } u=2x+4 \\ \text{ so we have } f(u)=6(\frac{u-4}{2})+3\]

Answer 4

\[f(u)=3(u-4)+3 \text{ since } \frac{6}{2}=3 \\ f(u)=3u-12+3 \text{ by distribution } \\ f(u)=3u-9\]

Answer 5

now finding f(5-3x) and f(5x-1) should be easy

Answer 6

just replace u in f(u)=3u-9 with 5-3x
just replace u in f(u)=3u-9 with 5x-1

Answer 7

Omg I see it! Thank you so much!

Answer 8

np

Answer 9

Yesss I got 6x - 6 which is the answer ^-^

Answer 10

cool i got that same thing