Problem: If $f(x^2 + 1) = x^4 + 5x^2 - 9$, then $f(x^2 - 3) = kx^4 + wx^2 + p$ where $k$, $w$, and $p$ are integers. Find the value of $(k + w + p)$.
I'm fine with doing problems where the argument is some expression and we have the original function $f(x)$, but how does one work in reverse to solve this problem (if my method is correct)? Otherwise, is there a way to skip directly from $f(x^2 + 1)$ to $f(x^2 - 3)$?
We have $$f(x^2+1) = x^4 + 5x^2 -9 = (x^2+1)^2 + 3(x^2+1) - 13$$ This means $$f(y) = y^2+3y-13$$ Now you should be able to complete it.
Hint:
Suppose $f(u)=u$, then if $f$ is the same exact function $f(x)=x$
Use this idea, and substitute $u^2-4=x^2$
Also for future reference suppose we have,
$$F(g(x))=z(x)$$
How do we get $F(x)$ back?
The standard procedure is to let $u=g(x)$. Then, $g^{-1}(u)=x$ (inverse function). Plugging everything in we get:
$$F(u)=z(g^{-1}(u))$$
So
$$F(x)=z(g^{-1}(x))$$
