MVT Differentiation Can anyone help me with these


*

*Suppose that $f$ is differentiable on $[4,6]$ and is such that $f(4) = 6$ and $f' \geq 2$ on
$[4,6]$. Find the lowest possible value of $f(6)$.


I think the answer for the lowest possible value of $f(6)$ is $10$. Is it right? 


*

*Give a function $f$ that is differentiable on $\Bbb R$ and which has the following 3 properties, or explain why such a function can't exist: $f(1) = 3$, $f(3) = 8$, and $f'(x)\leq 2$ for all $x$.


As per mean value theorem : $f'(c) = \frac{f(b)-f(a)}{b-a}$ where c is any point in the close interval [a b]
Therefore : 
$f' = \frac{f(6)-f(4)}{6-4} =2 $ 
$\Rightarrow  \frac{f(6)-6}{6-4} = 2$
$\Rightarrow = f(6)=6+4 =10$ 
