 # Theorem: Let f be continuous on [a,b] and assume f(a)!=f(b). Then for every lambda such that f(a)<lambda<f(b), there exists a c in (a,b) such that f(c)=lambda. Question: Suppose that f:[0,1]->[0,2] is continuous. Use the Intermediate Value Theorem to prove that their exists c in [0,1] such that: Leypoldon 2022-08-08 Answered
Theorem: Let $f$ be continuous on $\left[a,\phantom{\rule{thinmathspace}{0ex}}b\right]$ and assume $f\left(a\right)\ne f\left(b\right)$. Then for every $\lambda$ such that $f\left(a\right)<\lambda , there exists a $c\in \left(a,\phantom{\rule{thinmathspace}{0ex}}b\right)$ such that $f\left(c\right)=\lambda$.

Question:
Suppose that $f:\left[0,1\right]\to \left[0,2\right]$ is continuous. Use the Intermediate Value Theorem to prove that their exists $c\in \left[0,1\right]$ such that:
$f\left(c\right)=2{c}^{2}$

Attempt:
I know that when we have the condition were $f:\left[a,b\right]\to \left[a,b\right]$, the method to prove that c exits, is the same method you would use to prove the fixed point theorem.

Unfortunately I don't have an example in my notes when we have $f:\left[a,b\right]\to \left[a,y\right]$. How would I use the IVT to answer the original question?
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Hint:
Consider the function $g\left(x\right)=f\left(x\right)-2{x}^{2}$.
Some details:
$g\left(0\right)=f\left(0\right)\ge 0$ since the range of $f$ is contained in [0,2]. Similarly, $g\left(1\right)=f\left(1\right)-2\le 0$. Furthermore, $g$ is continuous since $f$ is. Now, either $g\left(0\right)$ or $g\left(1\right)==0$, and there's nothing to prove. Or $g\left(0\right)>0$, $g\left(1\right)<0$. The Intermediate value theorem assures there exists $c\in \left(0,1\right)$ such that $g\left(c\right)=0$.