Number of points where the function f(x) = te | vedclass

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(C) Let $g(x) = \sqrt{2x - x^2} = \sqrt{1 - (x-1)^2}$ and $h(x) = 2 - x$.We look for the intersection points: $\sqrt{2x - x^2} = 2 - x$.Squaring both

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$2$

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(C) Let $g(x) = \sqrt{2x - x^2} = \sqrt{1 - (x-1)^2}$ and $h(x) = 2 - x$.We look for the intersection points: $\sqrt{2x - x^2} = 2 - x$.Squaring both sides: $2x - x^2 = (2 - x)^2 = 4 - 4x + x^2$.$2x^2 - 6x + 4 = 0 \implies x^2 - 3x + 2 = 0 \implies (x-1)(x-2) = 0$.So,the curves intersect at $x = 1$ and $x = 2$.For $x  \sqrt{2x - x^2}$.For $1  2 - x$.For $x > 2$,$2 - x  2$ (domain is $[0, 2]$).At $x = 1$,the function changes from $h(x)$ to $g(x)$. The derivatives are $h'(1) = -1$ and $g'(x) = \frac{1-x}{\sqrt{2x-x^2}}$,so $g'(1) = 0$. Since $-1 
eq 0$,it is non-differentiable at $x = 1$.At $x = 2$,$g(2) = 0$ and $h(2) = 0$. The derivative $g'(x)$ as $x 	o 2^-$ is $\frac{1-2}{0^+} = -\infty$. Since the derivative is infinite,it is non-differentiable at $x = 2$.Thus,there are $2$ points of non-differentiability.

Number of points where the function $f(x) = \text{maximum}(\sqrt{2x - x^2}, 2 - x)$ is non-differentiable is:

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