Question

Please compute f(5), when f(2x+6)=3+x. Do I plug 5 into the 2x+6, or the 3+x first? Thanks!

Answer 1

Is the x in 3+x meant to be different?

Answer 2

The question says that y=f(x) is defined for all real numbers

Answer 3

Yes but if you plug 5 into that equation, f(x) does not equal 3 + x

Answer 4

yeah, I was thinking of setting 2x+6 equal to 5, so then x = -1/2, and then plugging that into 3+x

Answer 5

No, put 5 as x into the equation 2x+6, creating 16.

Answer 6

or should i plug 2x+6 into 3+x and get 9+2x?

Answer 7

Then put 16=x+3 and take 3 from both sides

Answer 8

ok, so then I add 16 +3 to get the answer of 19?

Answer 9

o, nevermind, so its 13

Answer 10

That's what I'd assume, although I'm not sure as I don't know why the x's would be different

Answer 11

yeah idk either. but dont you think the answer would be 19? cause it's f(16)=3+x

Answer 12

it's not 16=3+x

Answer 13

You should go with what you thought originally, that seems right. After re reading it, put 2x+6=5

Answer 14

hmm, ok, thanks

Answer 15

A decent way of doing so is to first solve \(2x + 6 = 5\) which comes out to be \(-\dfrac{1}{2}\) followed by plugging \(-\dfrac{1}{2}\) in. You will get \(3 + \left(-\dfrac{1}{2}\right)\)

Answer 16

ok, thank you!! That's how I decided to do it too

Answer 17

Geez, there's another way out too: The tricky one!

Answer 18

Notice how \(f(2x + 6) = x + 3 = \dfrac{1}{2}\left(2x + 6\right)\). One may conclude that \(f(x) = 0.5x\) and thus \(f(5) =\cdots\)

Answer 19

oh! That's smart. Thanks a bunch!