Find $f(7)$ when $f(x^2-2)=14|x|$. I was given this problem: 
If $f(x^2-2) = 14 |x|$, what is $f(7)$?
I'm not sure if i'm supposed to just plug $7$ into the equation. I was trying to get to the original $f(x)$, but I don't know how to. I tried to do it with a simpler function:
Let's say $f(x) = x^2$
If I plug in $x-1$ into the equation, then $f(x-1) = (x-1)^2$.
From here, I can't find any relation or operation to go from $f(x-1)$ to $f(x)$. Can someone help me? 
Thanks in advance :)
 A: $\bf Hint:$
Let $x=\pm\sqrt 9$.
${}{}{}$
A: $$x^2-2=7$$
$$x^2=9$$
$$x=\pm3$$
$$f(7)=14*|3|=42$$ 
$$f(7)=14*|-3|=42$$ 
A: Firstly, its worth noting that the given condition doesn't uniquely specify a function $f : \mathbb{R} \rightarrow \mathbb{R}$. In particular, since for all $x \in \mathbb{R}$ we have $x^2-2 \geq -2,$ hence the condition of interest only specifies the value of $f(u)$ when $u \geq -2$.
Nonetheless, since $7 \geq -2$, we can still work out the value of $f(7)$. It goes like this.
From $$\forall x \in \mathbb{R},f(x^2-2) = 14|x|,$$ we deduce $$\forall y \in \{\mathbb{R} \geq 0\}, f((\sqrt{y})^2-2) = 14 |\sqrt{y}|,$$ which simplifies to $$\forall y \in \{\mathbb{R} \geq 0\},f(y-2) = 14 \sqrt{y}.$$
Hence $$\forall u \in \{\mathbb{R} \geq -2\}, f((u+2)-2) = 14 \sqrt{u+2},$$ which simplifies to $$\forall u \in \{\mathbb{R} \geq -2\}, f(u) = 14 \sqrt{u+2}.$$
Hence
$$f(7) = 14\sqrt{9} = 42$$
A: I think it would be enlightening to do a step-by-step deconstruction of the given function and why the solution follows from certain definitions.
Some Definitions


*

*The preimage of a function is what you are "inputting" into your function.  These entities will be found as the parameter of the function, that is to say, what is inside the parenthesis after the $f$. In this case, the preimage of $f(7)$ is the number $7$. All the preimages of the given function will have the form $x^2-2$.

*The image of a function is what is "outputted" from the function. The image that will be produced from the given function will have the form $14\cdot |x|$. Images always have a corresponding input. Further, we can speak of the image of $7$ by denoting it $f(7)$.
Using the above definitions, we are asked to find the image $f(7)$ of the preimage $7$. From definition 1, I mentioned that all the preimages of the given function will have the form $x^2-2$, by which I mean that all the preimages will equal $x^2-2$. Therefore, to say that the preimage $7$ has the form $x^2-2$ is to write
$$x^2 - 2 = 7$$
from which it follows that $x = 3$ or $x = -3$. 
Therefore, when $x = 3$ or $x = -3$, we have the particular image $f(x^2-2) = 42$.
