Prove that $-2\leq f(2)-f(-1)\leq1$ Let $f:[-1,2]\to R$ be differentiable such that $0\leq f'(t)\leq 1$ for $t\in[-1,0]$ and $-1\leq f'(t)\leq 0$ for $t\in[0,2]$.Then prove that $-2\leq f(2)-f(-1)\leq1$.

Since $0\leq f'(t)\leq 1$ for $t\in[-1,0]$ and $-1\leq f'(t)\leq 0$ for $t\in[0,2]$
$\therefore k_1\leq f(t)\leq t $ for $t\in[-1,0]$ and $-t\leq f(t)\leq k_2$ for $t\in[0,2]$ where $k_1,k_2$ are constants.
Now i am stuck.How should i find $f(2)-f(-1)$?Please help me.
 A: Using Mean value theorem:   As $f$ is differentiable we may write 
$ f(2)-f(0)= 2 f'(c)$ for some $c \in [0,2 ]$, but $ -1 \leq f'(c) \leq 0 $, then $$ -2 \leq  f(2)-f(0) \leq 0 $$  Similarly, $ f(0)-f(-1)=  f'(c')$ for some $c' \in [-1,0 ]$, and so  $$ 0 \leq  f(0)-f(-1) \leq 1$$ Adding the two inequalities, we end up with $$ -2\leq  f(2)-f(-1) \leq 1$$
A: Hint: Use Newton - Leibniz formula and divide the integral in two parts : from $-1$ to $0$ and from $0$ to $2$. Finally use the monotonicity property of the definite integral to get the desired inequalities:
$$f(2)-f(-1)=\int\limits_{-1}^{2}{f'(x)dx}=\int\limits_{-1}^{0}{f'(x)dx}+\int\limits_{0}^{2}{f'(x)dx}\leq \int\limits_{-1}^{0}{1dx}+\int\limits_{0}^{1}{0dx}=1$$
and similarly
$$\int\limits_{-1}^{0}{f'(x)dx}+\int\limits_{0}^{2}{f'(x)dx}\ge \int\limits_{-1}^{0}{0 dx}+\int\limits_{0}^{2}{-1 dx}=-2$$
A: Hint:
From the FTC
$$f(2)-f(-1)=\int_{-1}^2f'(t)dt$$
and
\begin{align}
0\le\int_{-1}^0f'(t)dt\le 1 \qquad \qquad \int_{0}^2(-1)\,dt\le\int_{0}^2f'(t)dt\le \int_{0}^20\,dt
\end{align}
A: Note that
$$f(0) - f(-1) = \int_{-1}^0 f'(t) \, dt$$
$$f(2) - f(0) = \int_{0}^2 f'(t) \, dt$$
Since $0 \leq f'(t) \leq 1$ on $[-1, 0]$, we have
$$0 = \int_{-1}^0 0 \, dt \leq \int_{-1}^0 f'(t) \, dt \leq \int_{-1}^0 1 \, dt = 1$$
and therefore
$$0 \leq f(0) - f(-1) \leq 1$$
Similarly,
$$-2 = \int_{0}^2 -1 \, dt \leq \int_{0}^2 f'(t) \, dt \leq \int_{0}^2 0 \, dt = 0$$
and therefore
$$-2 \leq f(2) - f(1) \leq 0$$
Adding the two inequalities
$$ -2 = -2 + 0 \leq f(2) - f(-1) \leq 0 + 1 = 1 $$
