Prove that $m(y − x) ≤ f(y) − f(x) ≤ M(y − x)$

Question

Let $f : [a, b] \rightarrow \mathbb R$ be continuous and differentiable on $(a,b)$

For the derivative let $m \leq f′(z) \leq M$ for all $z \in (a,b)$ be valid.

How can I prove that $ m(y − x) ≤ f(y) − f(x) ≤ M(y − x)$ for all $x, y ∈ [a, b]$ with $x ≤ y$ ?

I thought of using the mean value theorem to find $f(y)-f(x) = (y-x)f'(z)$, some $z ∈ (x,y)$, but I neither know if this is a good start nor how to accomplish this.

Answer 1

Let's rewrite your inequality as $$m \leq \frac{f(y)-f(x)}{y-x} \leq M$$ We know by the mean value theorem that there exists a $z$ in $[a, b]$ such that $$f'(z) = \frac{f(y)-f(x)}{y-x}$$ Since $m \leq f'(z) \leq M$ the inequality is proved.

Answer 2

You are right indeed by MVT we have that

$$\frac{f(x)-f(y)}{x-y}=f’(c)$$

then use bounding limits for $f’(c)$.