h(x)=fog(x)=eg(x)−g(x)
=ex2−x−(x2−x)
h′(x)=ex2−x×(2x−1)−2x+1
=(2x−1)(ex2−x−1)
∵h(x) is increasing ∴h′(x)≥0 ⇒(2x−1)(ex2−x−1)≥0
Case 1: (2x−1)≥0 & x2−x≥0
\Rightarrow x\geq \frac{1}{2} & x\in (-\infty , 0]\cup [1,\infty )\Rightarrow x\in [1, \infty )
Case 2: (2x−1)≤0 and x2−x≤0
x<21 and x∈(0,1) ⇒x∈(0,21]
∴x∈[0,21]∪[1,∞)