Prove that if f is differentiable on $(a, b)$ and $f'$ is increasing, then $f$ is convex on $[a, b]$ 
A function $f$ is called convex on an interval $[a, b]$ if, for any $x, y \in
[a, b]$ and $t \in [0, 1]$ we have $f(tx + (1 − t)y) \leq tf(x) + (1 − t)f(y)$.

Would drawing a picture of this help in understanding this question conceptually and visually? The proof of this question requires the use of the mean value theorem. I am struggling to understand this real analysis question. 
 A: The image you should have is that if $f'$ is increasing then the tangent lines to the graph of $f$ will, if you look at points from left to right, will rotate counter-clockwisr, pointing ever higher and getting closer to become vertical. If $f$ was piecewise linear, then it would be shaped like a V, which is concave. This is actually the general idea.
Given $x<y$ and $t\in[0,1]$, the point $z=tx+(1-t)y$ is between $x$ and $y$, so the derivatives to the left of $z$ are smaller than the ones to the right. Use this with the mean value theorem on the intervals between $x$, $y$ and $z$. This use of the Mean Vale Theorem is a way to approximate $f$ by a piecewise linear map.
A: If $x < z < y$, then by the mean value theorem there exists $\xi_1,\xi_2$ such that $x < \xi_1 < z < \xi_2 < y$ and
$$\tag{1}\frac{f(z) - f(x)}{z - x} = f'(\xi_1) \leqslant f'(\xi_2) = \frac{f(y) - f(z)}{y-z}$$
Note that $\leqslant$ can be replaced with $<$ if $f'$ is strictly increasing (leading to strict convexity).
Since $z$ is between $x$ and $y$, we have $t \in (0,1)$ such that $z = tx + (1-t)y$. 
It follows that
$$\tag{2}z-x = (1-t)(y-x),$$ 
$$\tag{3} y-z = t(y-x)$$ 
I'll leave it to you to show using (1), (2) and (3) that
$$f(z) \leqslant tf(x) +(1-t)f(y),$$
proving convexity.
