Animated Solution for Mathematics - Functions: If $f:[1,\infty )\to [2,\infty )$ is given by $f(x)=x+1/x$, then $f^{-1}(x)$ equals

Understanding the Domain

Before we write a single equation, we must look at the domain. The problem defines our function on the interval $[1,\infty )$.

If we were to look at the function $f(x)=x+\frac{1}{x}$ over all real numbers, we would see a curve that dips and rises, failing the horizontal line test. It is not one-to-one.

By restricting the domain to $x\ge 1$, we are essentially taking the right-hand side of the graph, where the function is strictly increasing. At $x=1$, the function hits its minimum value of $f(1)=1+1=2$.

As $x$ grows, $f(x)$ grows. This monotonicity is the green light that allows us to find an inverse. Without this restriction, the inverse would not be a function at all.

Now, let us begin the dance. We want to find $f^{-1}(x)$. The standard operating procedure is to set $y=f(x)$.

So, we write:

Our goal is to isolate $x$. To clear the denominator, we multiply the entire equation by $x$:

Suddenly, the fog clears. We rearrange the terms to form a quadratic equation:

We have successfully transformed a rational function into a quadratic one. This is the power of algebraic manipulation—turning the unknown into the familiar.

The Tool of Choice

Since we cannot easily factor this, we reach for our most reliable tool: the quadratic formula. For any equation $a{x}^2+bx+c=0$, the roots are given by:

In our case, $a=1$, $b=-y$, and $c=1$. Substituting these values, we get:

This simplifies beautifully to:

We have two potential paths forward: the positive branch and the negative branch. This is the moment of truth. Which one is the true inverse?

Making the Choice

This is where many students stumble. We have two expressions, but only one is valid. We must return to our domain constraint: $x\ge 1$.

Let us test the negative branch: $x=\frac{y-\sqrt{y^2-4}}{2}$. If we pick a value for $y$, say $y=3$ (which is in our range $[2,\infty )$), the negative branch gives us:

Since $\sqrt{5}$ is approximately $2.236$, this results in $x\approx \frac{3-2.236}{2}=0.382$. But wait! $0.382$ is less than $1$.

This violates our domain constraint. The negative branch is an impostor. It does not belong to our function's domain. Therefore, we must reject it. The positive branch is the only one that respects the boundaries of our function.

With the negative branch discarded, we are left with $x=\frac{y+\sqrt{y^2-4}}{2}$. To express this as a function of $x$, we simply swap the variables.

The inverse function is:

Geometrically, this function is the reflection of our original curve across the line $y=x$. It is the 'undo' button for our original function.

You have navigated the domain constraints, mastered the quadratic transformation, and successfully identified the valid branch. This is not just solving a problem; this is mastering the logic of functions. Keep this rigor in your toolkit, and no problem will ever be too daunting.