MVT Differentiation Can anyone help me with these


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*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? 


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*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 $x$.


For the first one, you've done well in figuring out what the least value of $f(6)$ may be. By MVT, there is a point $4<t<6$ such that $$f'(t)=\frac{f(6)-f(4)}{6-4},$$ whence $$f(6)=f(4)+2f'(t)=6+2f'(t)\ge10.$$ A straight line shows us that $10$ is a feasible value.
For the second, use the same kind of reasoning to figure out the following:

Let $f$ be differentiable on $[1,3]$, such that $f(1)=3$ and $f'\le 2$ on $[1,3]$. What is the highest possible value of $f(3)$?

