Answer
See the explanation below.
Work Step by Step
Step 1. To prove $-1\leq f'(x)\leq1$, we need to show that $|f'(x)|\leq1$
Step 2. The given conditions indicate that we can use the Mean Value Theorem for the function.
Step 3. Assuming $x\lt w$, in the interval $(x,w)$, we can find a value $c$ such that $f'(c)=\frac{f(w)-f(x)}{x-w}$
Step 4. When $w\to x$, we have $c\to x$ because $c$ is sandwiched between $w$ and $x$; we have
$\lim_{w\to x}f'(c)=f'(x)$
Step 5. Taking the absolute value, we have $|f'(x)|=\lim_{w\to x}|f'(c)|=\lim_{w\to x}|\frac{f(w)-f(x)}{x-w}|$
Step 6. With the given condition that $|f(w)-f(x)|\leq|w-x|$ or $|\frac{f(w)-f(x)}{x-w}|\leq1$, the result from step 5 becomes $|f'(x)|\leq1$ which is what would be needed from step 1.
