If $f:[1,10] \to [1,10]$ is a non-decreasing function and $g:[1,10] \to [1,10]$ is non-increasing function. Let $h(x)=f(g(x))$ with $h(1)=1$, then $h(2)$:
(A) lies in $(1,2)$
(B)is more than $2$
(C) is equal to $1$
(D) is not defined
I know non-decreasing function means the slope is either $0$ or positive and non-increasing function means the slope is either $0$ or negative. I tried to solve this by assuming the function $f$ is increasing and $g$ is decreasing, but ended up at the wrong answer.
Kindly help me solve this problem.
Suppose $x_1<x_2$ then $g(x_1)\geq{}g(x_2) \implies f(g(x_1))\geq{}f(g(x_2))$ which proves that $h$ is decreasing. Now $2>1 \implies h(2)\leq{}h(1)=1$ and since $h(x)\in{}[1,10]$ we have $1\leq{}h(2)\leq{}1 \implies h(2)=1$.