Question 11

A function $f(x)$ is increasing on $[0, 1]$ if its derivative satisfies $f'(x) \ge 0$ everywhere on $[0, 1]$ (with $f'(x) > 0$ almost everywhere).

Analyze each function:

1. $f_1(x) = e^{-x} \implies f_1'(x) = -e^{-x}$. Since $e^{-x} > 0$ for all real $x$, $f_1'(x) < 0$ everywhere in $[0, 1]$. Thus, $f_1$ is strictly decreasing.
2. $f_2(x) = x^2 - \sin x \implies f_2'(x) = 2x - \cos x$. At $x = 0$, $f_2'(0) = 0 - \cos(0) = -1 < 0$. Therefore, $f_2$ is decreasing near $x = 0$ and is not increasing everywhere in $[0, 1]$.
3. $f_3(x) = \sqrt{x^3 + 1} \implies f_3'(x) = \frac{3x^2}{2\sqrt{x^3 + 1}}$. For $x \in [0, 1]$, $x^2 \ge 0$ and $\sqrt{x^3 + 1} \ge 1$, so $f_3'(x) \ge 0$ for all $x \in [0, 1]$ and $f_3'(x) > 0$ for all $x \in (0, 1]$. Hence, $f_3$ is increasing everywhere in $[0, 1]$.

Therefore, only function III is increasing everywhere in $[0, 1]$.