Algebra help

Question

Algebra help

algebranerd · Answer

Let f(x) = x2 − 81. Find f−1(x).

legomyego180 · Answer

$f(x)=x^2-81$ find $f^{-1}(x)$

Is this the question?

algebranerd · Answer

yes

algebranerd · Answer

anyone?

legomyego180 · Answer

So to solve for the inverse, everywhere you see a y (or in this case f(x)=y) substitute in an x, and every where you see a x subsittute in a y.

$$f(x)=x^2-81$$
$$x=y^2-81$$

Now that we have "flipflopped" our x's and y's we just simplify our equation so that we have a single y term.

$$x^2+81=y^2$$

Take the square root of both sides to get rid of the $y^2$

Which leaves you with $y=\sqrt{x^2+81}$ as your answer, but lets put it back in the terms the original question asked:

$$f^{-1}(x)=\sqrt{x^2+81}$$

legomyego180 · Answer

P.S. Dont be fooled by the square root and think that $$\sqrt{x^2+81}=x+9$$

You cant break up addition thats inside a square root like that:

$$\sqrt{a+b}\neq \sqrt{a}\sqrt{b}\neq \sqrt{a}+\sqrt{b}$$

mathstudent55 · Answer

@legomyego180
You did well, and you explained it well, but you missed something at the end.

In general, if

$x^2 = k$, then $x = \pm \sqrt k$

In your case, you have

$y^2 = x^2 + 81$

To solve for y, you get

$y = \pm\sqrt{x^2 + 81}$

The inverse function, then, is

$f^{-1} = \pm\sqrt{x^2 + 81}$

legomyego180 · Answer

you're right mathstudent. That totally slipped by me. Sorry about that AlgebraNerd