Prove there exists a $c$ such that $g(c) = 1$ in the interval $[-5, 5]$ I've this problem where I am asked to use the Intermediate Value Theorem:

A function $f$ is continuous where $f(-5) = -1$ and $f(5) = 6$, and $g(x) = 1 - (f(x))^2$.
Is there a value $c$ for $-5 \leq{c} \leq{5}$ such that $g(c) = 1$? Why, or why not?

I understand that the intermediate value theorem guarantees an $f(c) = 0$ so that $g(c) = 1 - 0^2$, but I cannot figure out a way to succinctly say there is a $g(c) = 1$ by this fact, since applying IVT to to $-5, 5$ yields $g(-5) = 0$ $g(5) = -35$, which does not guarantee a $g(c) = 1$.
Is there a theorem I am unaware of that proves this from the fact that there is an $f(c) = 0$?

Since $f$ is continuous on $[-5,5]$ and $f(-5)f(5) = -6 < 0$, by IVT there is a number $c \in (-5,5)$ such that $f(c) = 0 \implies f(c)^2 = 0 \implies g(c) = 1 - f(c)^2 = 1 - 0^2 = 1 - 0 = 1$.
