Intersection of a function having increasing derivative with tangent line I came across the following exercise in Spivak: 
Suppose that $f(x)$ is a differentiable function and suppose that $f'(x)$ is increasing. Show that every tangent line of $f(x)$ intersects the graph only once.

My intuition is as follows: 
Consider the tangent line $t(x)$ for any point $a$. Clearly, we have $t'(a) = f'(a)$. Since $f'(x)$ is an increasing function whereas $t'(x)$ is constant, the two graphs can never intersect again for $x\geq a$ because $f(x)$ "grows faster" than $t(x)$. 

How do make this argument precise?
Consider the tangent $l$ to $f(x)$ at $x=a$. This has gradient $f'(a)$. Let $c&gt;a$. Then by the mean value theorem, there is some $a&lt;b&lt;c$ such that $f'(b)(c-a)=f(c)-f(a)$. This means $f(c)=f(a)+f'(b)(c-a)&gt;f(a)+f'(a)(c-a)$ which is the value of $l$ at $c$. Do the same for the case $c&lt;a$.
Since $f'$ is increasing, the function $f$ is strictly convex, which means that its graph always lies "below the line segment joining any two points of the graph". 

Now assume that a tangent line $t$ at $f$ in the point $(x_1,f(x_1))$ intersects the graph of $f$ twice, at $(x_1,f(x_1))$ and $(x_2,f(x_2))$. Since the graph of $f$ has to lie below the two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$, the value of $(f-t)'$ cannot be zero at $x$, which is a contradiction, since $t$ should be a tangent at $f$ in $x_1$, i.e. $f'(x_1) = t'(x_1)$.