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(A) Given $f$ is an increasing function,so $f'(x) \geq 0$. Given $g$ is a decreasing function,so $g'(x) \leq 0$. We have $h(x) = f(g(x))$. Differentia

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(A) Given $f$ is an increasing function,so $f'(x) \geq 0$. Given $g$ is a decreasing function,so $g'(x) \leq 0$. We have $h(x) = f(g(x))$. Differentiating with respect to $x$,we get $h'(x) = f'(g(x)) \cdot g'(x)$. Since $f'(g(x)) \geq 0$ and $g'(x) \leq 0$,their product $h'(x) \leq 0$. This implies $h(x)$ is a decreasing function on $[0, \infty)$. Since $h(x)$ is decreasing,for $x \geq 0$,$h(x) \leq h(0)$. Given $h(0) = 0$,we have $h(x) \leq 0$. However,the codomain of $h$ is $[0, \infty)$,which means $h(x) \geq 0$ for all $x$. Combining $h(x) \leq 0$ and $h(x) \geq 0$,we conclude $h(x) = 0$ for all $x \in [0, \infty)$. Therefore,$h(x) - h(1) = 0 - 0 = 0$.

If $f$ and $g$ are increasing and decreasing functions respectively from $[0, \infty)$ to $[0, \infty)$,and $h(x) = f(g(x))$ with $h(0) = 0$,then $h(x) - h(1)$ is:

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