Let h(x) = min { x, x^2 } for every real numb | vedclass

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(D) Given $h(x) = \min \{ x, x^2 \}$.We compare $x$ and $x^2$:$x \le x^2 \Rightarrow x^2 - x \ge 0 \Rightarrow x(x - 1) \ge 0$.This inequality holds f

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All of the above

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(D) Given $h(x) = \min \{ x, x^2 \}$.We compare $x$ and $x^2$:$x \le x^2 \Rightarrow x^2 - x \ge 0 \Rightarrow x(x - 1) \ge 0$.This inequality holds for $x \le 0$ or $x \ge 1$.Thus,$h(x) = \begin{cases} x & x \le 0 \ x^2 & 0 $h(x)$ is continuous for all $x \in \mathbb{R}$ because the pieces match at $x=0$ $(0=0)$ and $x=1$ $(1=1)$.$h(x)$ is not differentiable at $x=0$ and $x=1$ because the left and right derivatives do not match at these points.For $x > 1$,$h(x) = x$,so $h'(x) = 1$.Therefore,all statements $(a)$,$(b)$,and $(c)$ are correct.

Let $h(x) = \min \{ x, x^2 \}$ for every real number $x$. Then:

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