Question

Can anyone verify this?

I'm finding the inverse of the function:

f(x)=x²+6x-5

SO far I got this as the inverse:

ƒ−1(x) = -3±√(x+14)

Here is the part i'm having trouble with, verifying the inverse:

ƒ−1(f(x))=x

for ƒ−1(f(x)), I was able to verify -3+√(x+14) = x:

ƒ−1(f(x))
=-3+√((x²+6x-5)+14)
=-3+√(x²+6x+9)
=-3+√(x+3)²
=-3+(x+3)
=x

However I'm having trouble in verifying
-3-√(x+14)=x

This is what I am ending up with:

ƒ−1(f(x))
=-3-√((x²+6x-5)+14)
=-3-√(x²+6x+9)
=-3-√(x+3)²
=-3-(x+3)
=-3-x-3
=-x-6

Can anyone help me please? Much appreciated, thanks

Answer 1

ƒ−1(f(x))
=-3-√((x²+6x-5)+14)
=-3-√(x²+6x+9)
=-3-√(x+3)²
=-3-√(-1)²(x+3)²
=-3-((-1)(x+3))
=-3-(-x-3)
=-3+x+3
=x

√9=√3²=√(1x3²)=√((-1)²x3²)
Since (-1)²=1

I just played with the numbers. I hope this helps.

Answer 2

hi firstly thanks for helping me, I'm just now looking at what you did there...

Answer 3

how did you get -1 out of (x+3)? In step 4?

Answer 4

if you were factoring -1 out of (x+3). wouldn't it become (-x-3)?

Answer 5

ƒ−1(f(x))
=-3-√(x+3)²
=-3-√1*(x+3)²
=-3-√(-1)²(x+3)²

(-1)²=1

Answer 6

* is the multiplication sign

Answer 7

Sorry, I forgot the braces.

ƒ−1(f(x))
=-3-√(x+3)²
=-3-√(1*(x+3)²)
=-3-√((-1)²(x+3)²)

Answer 8

ahh i see, so you defined 3² to be equal to (-1)²*(3)² and just substituted (-1)² into the square root then?

Answer 9

Exactly. ;)

Answer 10

Wow thank you very much, I never would have known that you could substitute something like under the square root!

Thanks again for the help.

Answer 11

like (-1)²*

Answer 12

You're welcome. :)

Answer 13

wait am i to delete this question now?

Answer 14

You can leave it here. No need to delete it. It may be a reference for persons having a similar mathematical problem. You can post another maths question if you want. :)

Answer 15

alrighty then, thanks.

Answer 16

:)