Suppose that $f:[0,1]→[0,2]$ is continuous then prove that there exists $c \in [0,1]$

such that $f(c)=2c^2$.

If a continuous function has values of an opposite sign inside an interval, then it has a root in that interval.

Let $f$ be the function satisfying the given hypothesis. Consider the function $g(x) = \sqrt{\frac{f(x)}{2}}$. Then $g$ is continuous function from $[0,1]$ to $[0,1]$ then by the following result, there exists $c \in [0,1]$ such that $g(c) = c$. i.e., $\sqrt{\frac{f(c)}{2}} = c$ and the proof is done.

If $f(0) = 0$ or $f(1) = 1$ then nothing to prove. We assume $f(0) \ne 0$ and $f(1) \ne 1$.

Consider $g(x) = f(x) - x$, then $g$ is a continuous function from $[0,1]$ to $[0,1]$.

Since $f$ is a function from $[0,1]$ to $[0,1$ we have $g(0) = f(0) - 0 > 0$ and $g(1) = f(1) - 1 < 0$. Now by intermediate value theorem there exists $x$ in $[0,1]$ such that $g(x) = 0 = f(x) - x$ and the proof is done.

**Proof :**

**Intermediate value theorem :**If a continuous function has values of an opposite sign inside an interval, then it has a root in that interval.

Let $f$ be the function satisfying the given hypothesis. Consider the function $g(x) = \sqrt{\frac{f(x)}{2}}$. Then $g$ is continuous function from $[0,1]$ to $[0,1]$ then by the following result, there exists $c \in [0,1]$ such that $g(c) = c$. i.e., $\sqrt{\frac{f(c)}{2}} = c$ and the proof is done.

**Result:**Let**$f : [0,1] \to [0,1]$ be a continuous function then there exists $c \in [0,1]$ such that f(c) = c.****Proof:**If $f(0) = 0$ or $f(1) = 1$ then nothing to prove. We assume $f(0) \ne 0$ and $f(1) \ne 1$.

Consider $g(x) = f(x) - x$, then $g$ is a continuous function from $[0,1]$ to $[0,1]$.

Since $f$ is a function from $[0,1]$ to $[0,1$ we have $g(0) = f(0) - 0 > 0$ and $g(1) = f(1) - 1 < 0$. Now by intermediate value theorem there exists $x$ in $[0,1]$ such that $g(x) = 0 = f(x) - x$ and the proof is done.

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