Finding the largest possible value of a non-defined function? 
I am confused, I've never had a similar problem. It seems pretty simple but I have no idea of the approach I should take. Hint me please. 
 A: The mean value theorem should be the way to approach this problem, since it relates the growth of $f$ to that of its derivative. Recall that the theorem says that if $f$ is continuous and differentiable, then 
 $$ \frac{f(a)-f(b)}{a-b}=f'(c) $$ for some $c$ between $a$ and $b$.
In this case, taking $a=2$, $b=0$, we have
 $$ \frac{f(2)-f(0)}{2-0}=\frac{f(2)+3}2=f'(c) $$
for some $c$ between $0$ and $2$. We do not know what $c$ is, but we know that $f'(c)\le 5$, regardless of the value of $c$. This gives us 
 $$ \frac{f(2)+3}2\le 5, $$
or $f(2)\le10-3=7$, so $7$ is the largest value $f$ could take. 
To see that this is indeed the answer (instead of a smaller number perhaps obtained by another method), we need an example where $f(2)=7$. Take $f(x)=-3+5x$. Then $f'(x)=5$ for all $x$. Also, $f(0)=-3$, and $f(2)=10-3=7$. 

Another user posted a solution which unfortunately does not work. Sadly, they deleted it. Since it is a natural approach, and the reason why it fails is subtle, I will indicate here what their approach was: They suggested to use the fundamental theorem of calculus, to get $$ f(2) = f(0) +\int_0^2 f'(x)\,dx, $$
so $f(2)\le f(0)+\int_0^25\,dx=10-3=7$. The problem is that the fundamental theorem does not apply in general to arbitrary derivatives and, in fact, there are derivatives that are not Riemann integrable. See here for an example. This would have been a correct argument if we knew, for example, that $f'$ is continuous.
A: The largest possible value will be when the function continously increases at a slope of $5$ , if it increases at any less value it wont be maximum , so it should be a straight line from $0$  to $2$ with a slope of 5 ie the value of  $f(2)$ will be $5x-3$ . Put $x = 2$ you will get $7$ . 
A: Think, at $x=0$, we know taht $f(0)=-3$. Then as we increase $x$, the function $f$ increases if $f'(x)>0$. If we want $f(2)$ to be as large as possible once we get to $x=2$, we should take $f'(x)$ to be as large as possible the 'whole way to' $x=2$. But the largest $f'(x)$ can be is $5$. So if we let $f'(x)=5$ from $x=0$ to $x=2$, then the total change in the function is 
$$
\Delta x\; f'(x)=(2-0)5=(2)5=10
$$
since we started at $f(0)=-3$, the largest possible value of $f(2)$ is $-3+10=7$.
A: The maximum is achieved when the derivative is at its maximum, that is, $f'(x)=5$. Thus $f(x)=5x+C$. Putting in $f(0) = -3$, we get $C=-3$. Thus $f(2)=7$.
