• Jan 26th 2009, 11:49 AM
william
$f(x)=2x-7$

Find:

I) $f(3x+2)$

II) $f^{-1}(x)$
• Jan 26th 2009, 12:20 PM
Mush
Quote:

Originally Posted by william
$f(x)=2x-7$

Find:

I) $f(3x+2)$

II) $f^{-1}(x)$

i) $f(3x+2) = 2(3x+2)-7$

ii) Your function takes x, multiples it by 2, and subtracts 7. So the inverse will divide by 2 and add 7!

$f^{-1}(x) = \frac{x}{2} + 7$
• Jan 26th 2009, 12:24 PM
william
Quote:

Originally Posted by Mush
i) $f(3x+2) = 2(3x+2)-7$

ii) Your function takes x, multiples it by 2, and subtracts 7. So the inverse will divide by 2 and add 7!

$f^{-1}(x) = \frac{x}{2} + 7$

Thank you very much sir. Will I) expand or is that the final result

Quote:

Originally Posted by Mush
i) $f(3x+2) = 2(3x+2)-7$

ii) Your function takes x, multiples it by 2, and subtracts 7. So the inverse will divide by 2 and add 7!

$f^{-1}(x) = \frac{x}{2} + 7$

will I) go to

f(3x+2)=6x+4-7

f(3x+2)=6x-3

??
• Feb 22nd 2009, 10:24 PM
mr fantastic
Quote:

Originally Posted by william
Thank you very much sir. Will I) expand or is that the final result

will I) go to

f(3x+2)=6x+4-7

f(3x+2)=6x-3

??

1. I don't know why you would use the word quadratics in the title of this thread since it is totally irrelevant to the question.

2. Does the question ask you to expand? Does anyone expect you to expand? Has the question been answered if you don't expand? Capisce?
• Feb 24th 2009, 03:38 PM
arpitagarwal82
Quote:

Originally Posted by Mush
i) $f(3x+2) = 2(3x+2)-7$

ii) Your function takes x, multiples it by 2, and subtracts 7. So the inverse will divide by 2 and add 7!

$f^{-1}(x) = \frac{x}{2} + 7$

I think the answer to the inverse will not be this. (second part of question).

While taking inverse, first add 7 and then divide whole thing by 2.

$y = 2x - 7$
=> $y + 7 = 2x$
$=> x = (y + 7)/2$

So inverse function is $f^-1(x) = (x +7)/2$