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?
 A: Consider the tangent $l$ to $f(x)$ at $x=a$. This has gradient $f'(a)$. Let $c>a$. Then by the mean value theorem, there is some $a<b<c$ such that $f'(b)(c-a)=f(c)-f(a)$. This means $f(c)=f(a)+f'(b)(c-a)>f(a)+f'(a)(c-a)$ which is the value of $l$ at $c$. Do the same for the case $c<a$.
A: Use the integral. We have, for any $x > a$:
\[
  f(x) = f(a) + \int_a^x f'(\xi)\, d\xi > t(a) + \int_{a}^{\frac{a+x}2} t'(\xi)\,d\xi = t(x) \]
(The latter holds as $f' \ge t'$ and $f' > t'$ on $[\frac{a+x}2, x]$. Argue analogously for $x < a$.
A: Assume that for some $c$, the tangent at $c$ cut the graph two times : at $c$ and at $d$. The slope of this line is $f'(c)$. However, according to MVT, there exists some $a \in (c,d)$ such that $f'(a)=\frac{f(d)-f(c)}{d-c}=f'(c)$. Since $a \neq c$, this contradicts the fact that $f'(x)$ is increasing.
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. $f'(x_1) = t'(x_1)$.
