Identify the correct statement about the func | vedclass

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(A) To analyze $f(x) = \max(x^2 - 1, 7 - x^2, 5)$,we find the intersection points of the curves:$1$. $x^2 - 1 = 7 - x^2 \implies 2x^2 = 8 \implies x^2

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$f(x)$ is not differentiable at $4$ points.

Vedclass · Answer

(A) To analyze $f(x) = \max(x^2 - 1, 7 - x^2, 5)$,we find the intersection points of the curves:$1$. $x^2 - 1 = 7 - x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$. At $x = \pm 2$,$f(x) = 3$.$2$. $x^2 - 1 = 5 \implies x^2 = 6 \implies x = \pm \sqrt{6}$.$3$. $7 - x^2 = 5 \implies x^2 = 2 \implies x = \pm \sqrt{2}$.Comparing the values,the function $f(x)$ is defined as:$f(x) = 7 - x^2$ for $x \in [-\sqrt{2}, \sqrt{2}]$$f(x) = 5$ for $x \in [-\sqrt{6}, -\sqrt{2}] \cup [\sqrt{2}, \sqrt{6}]$$f(x) = x^2 - 1$ for $x \in (-\infty, -\sqrt{6}] \cup [\sqrt{6}, \infty)$The function is continuous everywhere. The non-differentiable points occur at the intersection points $x = \pm \sqrt{2}$ and $x = \pm \sqrt{6}$,which are $4$ points.Thus,$f(x)$ is not differentiable at $4$ points. The correct option is $A$.

Identify the correct statement about the function $f(x) = \max(x^2 - 1, 7 - x^2, 5)$.

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