Approximating $f(1)$ given $f(0)=2$ and $\frac{1}{2}x^2≤f'(x)≤x^2$?

Question

This is an exercise using the mean value theorem:

Approximate $f(1)$ given $f(0)=2$ and $\frac{1}{2}x^2≤f'(x)≤x^2$ for $x≥0$.

I've found (using MVT):

$$\frac{1}{2}x^2+2≤f(1)≤x^2+2$$

and I can constrain this to

$$2≤f(1)≤x^2+2$$ by noting that $x≥0$ by assumption.

But what should I do with the upper bound?

Answer 1

Note that in mean value theorem, $$\frac{f(1)-f(0)}{1-0}=f'(x)$$ where $\large \color{red}{x \in (0,1)}$.

So we can re-state this as $$\frac{1}{2}x^2 \leq \frac{f(1)-f(0)}{1-0} \leq x^2$$ where $\large \color{red}{x \in (0,1)}$

Or better, $$\frac{1}{2}x^2+f(0) \leq f(1) \leq x^2+f(0)$$ where $\large \color{red}{x \in (0,1)}$

Hence, $$\frac{1}{2}\cdot (1)^2+f(0) \leq f(1) \leq 1^2+f(0)$$ Or,

$$\frac{5}{2} \leq f(1) \leq 3$$

Hope this helps.

Answer 2

By the Fundamental Theorem of Calculus $$ f(1)-f(0)=\int_0^1f'(t)\,\mathrm{d}t $$ Therefore, $$ \frac16\le f(1)-2\le\frac13 $$ That is, $$ \frac{13}6\le f(1)\le\frac73 $$ which is a bit tighter than the mean value theorem gives, if that matters.