Let f and g be increasing and decreasing func | vedclass

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(A) Given that $f$ is an increasing function and $g$ is a decreasing function on $[0, \infty)$.Let $x_1, x_2 \in [0, \infty)$ such that $x_1 < x_2$.Si

Vedclass · Accepted Answer

always zero

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(A) Given that $f$ is an increasing function and $g$ is a decreasing function on $[0, \infty)$.Let $x_1, x_2 \in [0, \infty)$ such that $x_1 Since $g$ is decreasing,$g(x_1) \geq g(x_2)$.Since $f$ is increasing,$f(g(x_1)) \geq f(g(x_2))$,which implies $h(x_1) \geq h(x_2)$.Thus,$h(x)$ is a decreasing function on $[0, \infty)$.Given $h(0) = 0$. Since $h$ is decreasing and its domain starts at $0$,for any $x \geq 0$,$h(x) \leq h(0) = 0$.Also,the range of $f$ and $g$ is $[0, \infty)$,so $h(x) = f(g(x)) \geq 0$.Therefore,$h(x) = 0$ for all $x \geq 0$.Consequently,$h(x) - h(1) = 0 - 0 = 0$.

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

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