Let f(x) = 15 - |x - 10|; x in R. Then the se | vedclass

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(A) Given $f(x) = 15 - |x - 10|$.We need to find the points where $g(x) = f(f(x))$ is not differentiable.$g(x) = f(f(x)) = 15 - |f(x) - 10| = 15 - |(1

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$\{5, 10, 15\}$

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(A) Given $f(x) = 15 - |x - 10|$.We need to find the points where $g(x) = f(f(x))$ is not differentiable.$g(x) = f(f(x)) = 15 - |f(x) - 10| = 15 - |(15 - |x - 10|) - 10| = 15 - |5 - |x - 10||$.The function $f(x)$ is not differentiable at $x = 10$ (the vertex of the absolute value function).The composite function $g(x) = f(f(x))$ is not differentiable at points where $f(x)$ is not differentiable,or where the inner function $f(x)$ takes the value $10$ (the point where the outer $f$ is not differentiable).$1$. $f(x)$ is not differentiable at $x = 10$.$2$. $f(x) = 10 \implies 15 - |x - 10| = 10 \implies |x - 10| = 5 \implies x - 10 = 5$ or $x - 10 = -5 \implies x = 15$ or $x = 5$.Thus,the set of points where $g(x)$ is not differentiable is $\{5, 10, 15\}$.

Let $f(x) = 15 - |x - 10|; x \in R$. Then the set of all values of $x$,at which the function $g(x) = f(f(x))$ is not differentiable,is

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