Let $f:[a,b]\rightarrow \mathbb{R}$ a continuous function, with continuous derivate in $[a,b]$ such that:

Question

$0<f'(x)<M \ \ \forall \ x\in[a,b]$

Find $c,d\in R$ such that $c\leq f(x)\leq d$.

Answer 1

For extreme value theorem the function $f$ admits at least a maximum and a minimum on $[a,b]$. Since $f'>0$, the function is strictly increasing and takes extreme values at the bounds of the interval of definition. Then $c=f(a)$ and $d=f(b)$.

Answer 2

hint

$f'$ is continuous at $[a,b[$, by integration

$$(\forall x\in[a,b]) \;\; 0<f'(x)<M\implies$$ $$0 \leq\int_a^xf'(t)dt<M\int_a^xdt \implies$$ $$0 \leq f(x)-f(a)<M(x-a)\implies$$ $$f(a) \leq f(x)<M(b-a)+f(a)$$

Answer 3

If you know a bit of general topology, there is an easy proof of a stronger statement:

Lemma. Let $(X, \tau), (Y, \sigma)$ be topological spaces and let $f \colon (X, \tau) \to (Y, \sigma)$ be continuous. If $C$ is compact in $(X,\tau)$. Then $f[C]$ is compact in $(Y, \sigma)$.

Prove this and combine it with the fact that $C \subseteq \mathbb{R}$ is compact in $\mathbb{R}$ if and only if it is closed and bounded.