# Intermediate Value Theorem Question to show another function:

How would I go about showing this question:

Suppose that the function $$f:[0,1] \rightarrow [0,1]$$ is continuous. Use the Intermediate Value Theorem to prove that there exists $$c \in [0,1]$$ such that $$f(c) = c(2-c^2)$$

My thoughts are we can define a function say $$g(x) := f(x) - x(2-x^2)$$ and then $$g(0) = f(0)$$and then $$g(1) = f(1) - 1$$ and then I'm stuck on how to complete it to get to the point where

$$g(0) < 0 < g(1)$$ so there would exist a c s.t. $$g(c) = 0$$ [by IVT] and thus $$f(c) = c(2-c^2)$$

If anyone could help, would be much appreciated thank you!

Hint: Try to prove instead that $$g(0) \ge 0 \ge g(1)$$.
• @Ian Adding in equality doesn't make things any more difficult. If we get $g(0) = 0$ or $g(1) = 0$, then it just means that our choice of $c$ will be $0$ or $1$. Mar 7, 2021 at 16:16