Can inverse of every invertible function $f: \mathbb{R} \to \mathbb{R}$ be drawn by reflecting that function on $y = x$? Let $f: \mathbb{R} \to \mathbb{R}$ be an invertible function.
Can the graph of $f^{-1}$ be obtained by reflecting the graph of $f$ on the line $y = x$?
If yes then why?
 A: Reflecting the graph of a function on the line $y=x$ in order to obtain the graph of the inverse function is bad practice, because it mixes up the names of the variables. As long as you are talking about some $f$ and its inverse $f^{-1}$ at the same time you should draw just one curve. This curve then can be viewed as graph of the function $x\mapsto y=f(x)$ and at the same time as graph of the inverse function $y\mapsto x=f^{-1}(y)$, whereby you have to tilt your head $90^\circ$ in order to see the graph of $f^{-1}$ over a horizontal axis (directed to the left).
When you are through with studying how the properties of $f$ (e.g., the derivative of $\exp$) are reflected in the properties of $f^{-1}$ ($\log$ in this case) you are free to draw a standard picture of the graph of $f^{-1}$ on a second piece of paper. The resulting curve will then indeed be a mirror image of the original curve, as indicated in your question.
A: Yes, because by definition $x=f^{-1}(y) \Leftrightarrow y=f(x)$, so $f$ "becomes" $f^{-1}$ if you change the axes.
A: Consider a function $y=f(x)$
If $f(x)$ is invertible, you swap $y$ and $x$ to invert it and  the function will change to $x=f(y)$. Since the old $y$ is the new $x$ and vice versa, this is exactly the same as reflecting it in the line $y=x$
