The function defined by f(x) = max {x^2, (x - | vedclass

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(D) Let $g_1(x) = x^2$,$g_2(x) = (x - 1)^2$,and $g_3(x) = 2x(1 - x) = 2x - 2x^2$. We analyze the intersections in the interval $[0, 1]$: $1$. $g_1(x)

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is not differentiable at more than two points.

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(D) Let $g_1(x) = x^2$,$g_2(x) = (x - 1)^2$,and $g_3(x) = 2x(1 - x) = 2x - 2x^2$. We analyze the intersections in the interval $[0, 1]$: $1$. $g_1(x) = g_2(x) \implies x^2 = x^2 - 2x + 1 \implies 2x = 1 \implies x = 1/2$. $2$. $g_1(x) = g_3(x) \implies x^2 = 2x - 2x^2 \implies 3x^2 - 2x = 0 \implies x(3x - 2) = 0 \implies x = 2/3$ (since $x \in [0, 1]$). $3$. $g_2(x) = g_3(x) \implies x^2 - 2x + 1 = 2x - 2x^2 \implies 3x^2 - 4x + 1 = 0 \implies (3x - 1)(x - 1) = 0 \implies x = 1/3$ or $x = 1$. Comparing the values in the intervals: - For $x \in [0, 1/3]$,$g_3(x)$ is the maximum. - For $x \in [1/3, 1/2]$,$g_2(x)$ is the maximum. - For $x \in [1/2, 2/3]$,$g_1(x)$ is the maximum. - For $x \in [2/3, 1]$,$g_3(x)$ is the maximum. The function $f(x)$ is non-differentiable at the points where the maximum switches: $x = 1/3$,$x = 1/2$,and $x = 2/3$. Thus,the function is non-differentiable at three points.

The function defined by $f(x) = \max \{x^2, (x - 1)^2, 2x(1 - x)\}$ for $0 \le x \le 1$:

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