Spivak's Calculus, Ch. 11, "Significance of the Derivative", prob. 58: Prove $f'$ increasing then every tangent line intersects graph of $f$ only once The following is a problem from ch. 10, "Significance of the Derivative", from Spivak's Calculus



*Prove that if $f'$ is increasing, then every tangent line of $f$ intersects the graph of $f$ only once.


My proof seems at first glance to be the same as the solution manual's proof, but I don't understand the last step in the terse solution manual proof.
Here is my proof
$f'$ increasing means $\forall x \forall y, x<y \implies f'(x)<f'(y)$.
Consider a point $(x_0, f(x_0))$ on the graph of $f$.
The tangent line at this point has slope $f'(x_0)$.
Assume this line intersects the graph of $f$ at $(x,f(x))$.
Then $f'(x_0)=\frac{f(x)-f(x_0)}{x-x_0}$.
By the Mean Value Theorem we know that
$$\exists c, c \in (x_0, x) \land f'(c)=\frac{f(x)-f(x_0)}{x-x_0}$$
$$\implies f'(x_0)=f'(c)$$
$$\bot$$
The contradiction occurs because $f'$ is increasing by assumption.
Therefore, by proof by contradiction, we conclude that the tangent line line at any point does not intersect the graph of $f$ at any other point.
Here is the proof from the solution manual
The tangent line through $(a,f(a))$ is the graph of
$$g(x)=f'(a)(x-a)+f(a)$$
$$=f'(a)x+f(a)-af'(a)$$
If $g(x_0)=f(x_0)$ for some $x_0 \neq a$, then
$$0=g'(x)-f'(x)=f'(a)-f'(x), \text{ for some } x \text{ in } (a,x_0) \text{ or } (x_0,a)\tag{1}$$
This is impossible, since $f'$ is increasing.
Where does $(1)$ come from?
 A: It is Rolle's theorem/mean value theorem, applied to the function $h(x) = g(x) - f(x)$.  Under the assumption that $h(a) = h(x_0)$ for distinct numbers $a$ and $x_0$, there must be a number $x$ between $a$ and $x_0$ for which $h'(x) = 0$.
A: We have that $g(a)=f(a)$ and $g(x_0)=f(x_0)$ with $a\neq x_0$. This means $h(x)= g(x)-f(x)$ is zero at $x=a, x_0$. Clearly $h$ has a derivative (f is differentiable and g is a lone so also differentiable). If a differentiable function is equal at two points, there is a point in the middle where $h’(x)=0$ (Mean value theorem shows this if you like). This gives the first part of the statement.
Now notice that $g’(x)=f’(a)$ for all $x$ as $g$ is the tangent line to $f$ at $a$. This gives the second part by replacing $g’(x)$ with $f’(a)$. This contradicts that $f’$ is increasing.
As a note, I think there is an issue with your proof. The first line is the definition of $f$ being increasing, not $f’$. These aren’t equivalent, for example $f(x)=x^2$ is not increasing but it’s derivative is. I also don’t see where you get your formula for $f’(x_0)$. Hope this helps.
A: I much prefer your proof (after my edit for typos). From the definition of $g$ we have $g'(x)=f'(a)$ for all $x,$ so $g'(x)-f'(x)=f'(a)-f'(x)$ for all $x.$ So the function $h(x)=g(x)-f(x)$ has non-$0$ derivative for $x\ne a.$  At this point we can either say (i) the function $h(x)=g(x)-f(x)$ has negative derivative for $x>a$ so $h$ is decreasing for $x>a$ so $x>a\implies g(x)-f(x)=h(x)<h(a)=0,$ & a similar argument shows $x<a\implies g(x)-f(x)>0,$ OR (ii) use Rolle's Theorem: If $h(x)=h(a)=0$ for some $x\ne a$ then for some $y$ strictly between $x$ and $a$ we have $0=h'(y)$.
