Question

If f(x) = x + (1/x); x <= -1, f^-1(x)= ?

Answer 1

f^{-1}(x)=\frac{x-\sqrt{x^2-4}}{2} \quad \text{for } x\le -2.

Answer 2

Solution:

We are given

$$f(x)= x+\frac{1}{x}\quad \text{with } x\le -1.$$

Let

$$y=x+\frac{1}{x}.$$

Multiply both sides by $x$ (note $x\neq0$):

$$yx=x^2+1.$$

Rearranging, we obtain the quadratic in $x$:

$$x^2-yx+1=0.$$

Using the quadratic formula,

$$x=\frac{y\pm\sqrt{y^2-4}}{2}.$$

Since the original domain is $x\le -1$ (and the range of $f$ is $y\le -2$), we choose the branch that gives $x\le -1$. Testing, for example, with $y=-3$:

$$\frac{-3+\sqrt{9-4}}{2}\approx \frac{-3+2.236}{2}\approx -0.382 \quad (\text{not in domain})$$ $$\frac{-3-\sqrt{9-4}}{2}\approx \frac{-3-2.236}{2}\approx -2.618 \quad (\text{valid}).$$

Thus, the inverse function is:

$$f^{-1}(y)=\frac{y-\sqrt{y^2-4}}{2}\quad \text{for } y\le -2.$$

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Explanation (Minimal):

1. Write $y = x + \frac{1}{x}$.
2. Multiply by $x$ to get $x^2 -yx + 1=0$.
3. Solve using the quadratic formula.
4. Select the branch $\frac{y-\sqrt{y^2-4}}{2}$ to satisfy $x\le -1$.