Let f : [0,1] to [0,1] be a continuous functi | vedclass

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(C) Define a new function $g(x) = f(x) - x$ for $x \in [0, 1]$.Since $f : [0, 1] 	o [0, 1]$,we have $0 \leq f(x) \leq 1$ for all $x \in [0, 1]$.At $x

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must have atleast one solution in $[0,1]$

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(C) Define a new function $g(x) = f(x) - x$ for $x \in [0, 1]$.Since $f : [0, 1] 	o [0, 1]$,we have $0 \leq f(x) \leq 1$ for all $x \in [0, 1]$.At $x = 0$,$g(0) = f(0) - 0 = f(0) \geq 0$.At $x = 1$,$g(1) = f(1) - 1 \leq 0$ (since $f(1) \leq 1$).If $g(0) = 0$ or $g(1) = 0$,then $x = 0$ or $x = 1$ is a solution to $f(x) = x$.If $g(0) > 0$ and $g(1) Therefore,the equation $f(x) = x$ must have at least one solution in $[0, 1]$.

Let $f : [0,1] \to [0,1]$ be a continuous function,then the equation $f(x) = x$

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