Find maximum of $f(0)$

Question

Let $f$ be a continuously differentiable function on $(-1,4)$, $f(3)=5$ and $f'(x)\geq-1$ for all $x$. Find maximum of $f(0)$.

I have plotted the graph of $f$ and found the maximum is $8$. But I didn't find the solution exactly.

Answer 1

Since $f$ is continuously differentiable the followng quantiy is well defined : $$\int_0^3f'(x)dx$$ Firstly knowing that $f(3)=5$ we get : $$\int_0^3f'(x)dx=f(3)-f(0)=5-f(0)$$ And since $f'(x)\geq-1$ : $$\int_0^3f'(x)dx\geq-3$$ Finally : $$5-f(0)\geq-3 \iff f(0)\leq8$$

Answer 2

By MVT

$$f (0)=f (3)-3f'(c)=5-3f'(c) $$

with $c\in (0,3) $.

but $f'(c)\geq -1$ thus

$$f (0)\leq 5+3$$

Answer 3

$f(0) = f(3) - \displaystyle \int_0^3 f'(x)\,dx$.

$f(3)-f(0) = \displaystyle \int_0^3 f'(x) \,dx$

$f(3)-f(0) \geq -3$

$5 - f(0) \geq -3$.

$8 \geq f(0)$.