Let $ f(x) $ be a differentiable function in $ \mathbb{R} $ such that $ f'(x) $ is injective. Let $ x_0 \in \mathbb{R} $ and $ L $ be the line tangent to $ f $ at the point $ x_0 $. I want to show that $ L $ does not intersect $ \{(x, f(x) \mid x \in \mathbb{R}\} $ anywhere else. Is my proof valid?
Every line in the plane that is not perpendicular to the $ y $-axis can be described as a linear function. $ L $ is a line in the plane. Therefore, there exists a linear function $ l(x) = ax + b$ that aligns completely with $ L $. $ a $ is the slope of the line tangent to the graph of $ f $ at $ x_0 $, which means $ a = f'(x_0) $.
Suppose, for the sake of contradiction, that there exists $ t_0 \neq x_0 $ such that $ L $ intersects $ f $ at $ t_0 $, and WLOG, assume $ t_0 \gt x_0 $. Therefore, $ l(t_0) = f(t_0) $.
If a function is differentiable in an interval $ I $, then it is continuous in $ I $. $ f $ is differentiable in $ \mathbb{R} $. Therefore, $ f $ is continuous in $ \mathbb{R} $. If a function is continuous/differentiable in an interval $ I $, and $ J \subseteq I $ is an interval, then the function is continuous/differentiable in $ J $. Therefore, $ f $ is continuous in $ [x_0, t_0] $ and differentiable in $ (x_0, t_0) $.
From the Mean Value Theorem, we can deduce that there exists $ c \in (x_0, t_0) $ such that $ f'(c) = \frac{f(t_0)-f(x_0)}{t_0-x_0} $. But, $ \frac{f(t_0)-f(x_0)}{t_0-x_0} = \frac{l(t_0)-l(x_0)}{t_0-x_0} = a = f'(x_0) $.
$ x_0 \neq c $ and $ f'(x_0) = f'(c) $, which means $ f'(x) $ is not injective - contradiction.