Answer
See explanations.
Work Step by Step
Step 1. Assume the function $f(x)$ satisfies the conditions required by the Mean Value Theorem; we have point $c$ in the interval of $(a, b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$
Step 2. First prove $f(-1)\lt f(1)$, as $f'(x)\gt0$, the function increases as $x$ increases; thus we have $f(1)\gt f(-1)$
Step 3. To prove that $f(1)\lt 2+f(-1)$, rewrite the inequality as $f(1)-f(-1)\lt 2$. From step 1, let $a=-1, b=1$; we have $f'(c)=\frac{f(1)-f(-1)}{1+1}$ which gives $f(1)-f(-1)=2f'(c)$. As the condition $f'(x)\lt 1/2$ is true for $x=c$, we have $f(1)-f(-1)=2f'(c)\lt 1\lt 2$
Step 4. Combining the above results, we have $f(-1)\lt f(1)\lt 2+f(-1)$
