Question

For the function f(x)=(7-8x)^2, find f^-1. Determine whether f^-1 is a function.

Answer 1

Little help guys, I'm not sure how to do things like this.

Answer 2

what exactly is the questoin asking

Answer 3

It's asking to solve that equation for f^-1, and determine whether it's a function or not.

Answer 4

the inverse is not a function because
\[f(x)=(7-8x)^2\] is not a one to one function

Answer 5

okay. Can you explain how to find that out?

Answer 6

the clue is the square

Answer 7

if you graph the function you will get a parabola, which fails the "vertical line test" for sure

Answer 8

oops i meant "horizontal line test"

Answer 9

Okay. ALSO, do you know how to solve it to find f^-1?

Answer 10

alternatively you can try to solve
\[x=(7-8y)^2\] for \(y\)
if you do it carefully you will see that the inverse in not a function

Answer 11

most common mistake is to solve like this
\[x=(7-8y)^2\\
\sqrt{x}=7-8y\\
sqrt{x}-7=-8y\\
\frac{\sqrt{x}-7}{-8}=y\] but that is wrong

the first step should be
\[\pm\sqrt{x}=7-8y\\
\pm\sqrt{x}-7=-8y\\
\frac{\pm\sqrt{x}-7}{-8}=y\]

Answer 12

and it is the \(\pm\) that shows the inverse is not a function

Answer 13

So f^-1(x)=(7+-(sqrt x))/8?

Answer 14

yes

Answer 15

okay, and it's not a function?

Answer 16

although it is an abuse of notation to write
\[f^{-1}(x)=\frac{\pm\sqrt{x}-7}{8}\] because when you write \(f^{-1}\) you are saying it is a function, even though it is not
yes, it is not a function

Answer 17

Can you help me with another question?

Answer 18

go ahead and post, i will look at it

Answer 19

g(x)=14/(x+3); Find (g^-1*o*g)(4)