# Find $f(y^2)$ given $f(x)$ [closed]

This seems sort of like an inverse function but I'm not sure how to find the final answer. Do I just find the inverse of $$F(x)$$ and then square it?

$$F(x) = \frac{1}{x-8}$$.

Simplify $$F(y^2)$$

• Welcome to MSE. You have to surround the MathJax in $ signs for the formatting command to work. Sep 8, 2019 at 16:45 • No, there is no inverse involved. Can you do$F(12)$? Can you do$F(932)$? If so, try doing$F(y^2)$. Sep 8, 2019 at 16:46 • To answer the question, just write$y^2$where you see$x$; don’t let it confuse you that$y$sometimes represents$f(x)$; in this question,$y^2\$ is the argument of the function Sep 8, 2019 at 16:47
• Why isn't it just $${1\over y^2-8}?$$ Sep 8, 2019 at 16:47
• Okay I just use y^2 for the input that makes sense. I was over-complicating it. Thank you everyone for your help! I reformatted the exponent but I could get the fraction to format what did you do for that @saulspatz? Thanks again for the help! Sep 8, 2019 at 16:55

A function is like a machine that takes an input and gives an output. The specific rules that the machine follows in order to determine the output used are presented in various ways, one of the options being how it is presented here.

$$F(\color{red}x)=\dfrac{1}{\color{red}x-8}$$

This means, that to get our output, we take our input (in this case $$x$$), and then subtract eight from it and divide 1 by the result.

We now know how to find, say, $$F(\color{red}2)$$ as being $$F(\color{red}2)=\dfrac{1}{\color{red}2-8}$$

This is the same regardless of how complicated or simple of an input we desire, be it a complicated number or even an algebraic expression.

$$F(\color{red}{5.88876771}) = \dfrac{1}{\color{red}{5.88876771}-8}$$

$$F(\color{red}{x^2+5}) = \dfrac{1}{\color{red}{x^2+5}-8}$$

Or in your case, even using a variable name different than $$x$$:

$$F(\color{red}{y^2}) = \dfrac{1}{\color{red}{y^2}-8}$$

That should be pretty straight forward. To go from f(x) to $$f(y^2)$$, replace every "x" by "$$y^2$$". Here, $$f(x)= \frac{1}{x- 8}$$ so $$f(y^2)= \frac{1}{y^2- 8}$$.