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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!

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1 Answer 1

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Hint: Try to prove instead that $g(0) \ge 0 \ge g(1)$.

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  • $\begingroup$ I didn't know you could apply IVT via that inequality-thanks! $\endgroup$
    – user887539
    Mar 7, 2021 at 16:13
  • $\begingroup$ @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 $1$. $\endgroup$ Mar 7, 2021 at 16:16

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