Graphs of functions

Question

Is there a rule or way that will help us graph $f(\frac{1}{x})$ if $f(x)$ graph is known? Or is the only way plotting values? I thought of this question when we learned that the graph $f^{-1}(x)$ is the mirror image of $f(x)$ across $y=x$.

Answer 1

That's a good question, but the answer is not so satisfying. In the situation of an inverse function, $y = f^{-1}(x)$ if and only if $x = f(y)$. So, naturally, you're switching the $x$ and $y$ when passing from the graph of $f$ to the graph of $f^{-1}$. As you point out, this is a reflection across the line $y=x$, which is a particularly nice type of linear transformation. In fact, it's an isometry (a.k.a. congruence, a.k.a. rigid motion) of the plane, which means that it preserves distances and angles (and the shape and size of objects).

Unfortunately, the passage from $y = f(x)$ to $y = g(x) = f(\frac{1}{x})$ is not such a well-behaved transformation. What you have to do is understand $$ x \,\longleftrightarrow\, \frac{1}{x}. $$ For simplicity, let's restrict our attention to $x>0$. A point $(x, y)$ is on the graph of $y = f(x)$ if and only if the point $(\frac{1}{x}, y)$ is on the graph of $y = g(x)$. Replacing a number by its reciprocal distorts distances: for instance, the distance between $1$ and $2$ is the same as the distance between $10$ and $11$, but for their reciprocals, the distance between $1$ and $\frac{1}{2}$ is much greater than the distance between $\frac{1}{10}$ and $\frac{1}{11}$. The infinite ray $[1, \infty)$ is flipped over and contracted to cover the interval $(0, 1]$ by taking reciprocals.

The behavior of $f(x)$ as $x \to 0$ determines the behavior of $g(x)$ as $x \to \infty$ and vice versa, but the shape of the curve changes.