Question

For the function f(x)= (8-2x)^2 , find f^-1. determine whether f^-1function

Answer 1

solve it using limit

Answer 2

Inverse...? \(\large\frac{1}{f(x)}\)

Answer 3

Or do you mean the anti-function, as in anti-derivative for example?

Answer 4

He is saying inverse function
y=(8-2x)^2
x=(8-2y)^2
(+/-)(sqrt(x)-8)/2=y
The inverse is not a function because there are 2 possible values for y

Answer 5

FYI about being careful about notation:

\[\tan^{-1} \theta = arctan \theta \neq \frac{1}{\tan \theta} = \cot \theta\]

\[(x^2+1)^{-1} = \frac{1}{x^2+1}\]

Answer 6

@ivan514 , If you assume that, then yes. I don't presume what he/she intended. :-)

Answer 7

Statistically, most questions here are algebra2 level.. So most likely thats what he asked

Answer 8

question is to find the inverse
but it is not posed correctly because you cannot say
"find the inverse function and then say if it is a function"

Answer 9

this function is not one to one and so it does not have an inverse

Answer 10

@ivan514 Meh... that's a bad assumption to make, it depends on the day and who's asking. As they say, "There's lies, damn lies, and statistics." lol

I agree with @satellite73

Answer 11

you can write
\[x=(8-2y)^2\]
\[\pm x=8-2y\]
\[2y=8\pm\sqrt{x}\]
\[y=4\pm\frac{\sqrt{x}}{2}\] if you like, but that does not make y a function of x

Answer 12

typo on line two should be
\[\pm\sqrt{x}=8-2y\]

Answer 13

I understand its not a function but i its giving me 2 answer choices
+/- = 8+/- sqrt x/2 and +/- sqrt 8+x /2

Answer 14

Helpful syntax display tip:

\pm \(\rightarrow\ \pm\)

Put a \( before the \pm and

Answer 15

a \) after it to activate

Answer 16

\(8 \pm \sqrt{\large\frac{x}{2}}\) and \(\pm\sqrt{8 + \large\frac{x}{2}}\) , yes?

Answer 17

Yes . These are my 2 answer choice.

Answer 18

What I had:

\[y=(8-2x)^2\]
\[\pm \sqrt{y}=8-2x\]
\[\pm \sqrt y - 8 = -2x\]
\[\frac{\pm\sqrt{y}-8}{-2}=x\]
\[\frac{8\pm\sqrt{y}}{2}=x\]
\[4\pm \frac{\sqrt{y}}{2}=x\]
switching x & y to make the switched around, anti-function :
(the function that undoes what f(x) does)
\[4\pm \frac{\sqrt{x}}{2}=y\]


Same thing as @satellite73 had, he just switched the independent & dependent variables at the start, whereas I did it at the end. Are you sure it's not: f(x)=(16-2x)^2 ???

Answer 19

This is also valid:
\[\frac{8\pm \sqrt{x}}{2}=y\]

But:
\[\frac{8\pm \sqrt{x}}{2} \neq 8\pm \frac{\sqrt{x}}{2} \neq \pm\sqrt{8 + \frac{x}{2}}\]

Answer 20

I final1y got the answer. Thanks so much you guys !