Math Snap

STEP 1

1. The function $f(x) = \sqrt{x^2 + 2x}$ is defined for $x > 0$.
2. The inverse function, $f^{-1}(x)$, will satisfy $f(f^{-1}(x)) = x$.
3. We need to solve for $x$ in terms of $y$ where $y = f(x)$.

STEP 2

1. Set up the equation for the inverse.
2. Solve for $x$ in terms of $y$.
3. Verify the inverse function.

STEP 3

Set up the equation for the inverse by letting $y = f(x)$. Thus, we have:
$$y = \sqrt{x^2 + 2x}$$

STEP 4

To solve for $x$ in terms of $y$, first square both sides to eliminate the square root:
$$y^2 = x^2 + 2x$$ Rearrange the equation to form a quadratic equation:
$$x^2 + 2x - y^2 = 0$$

STEP 5

Solve the quadratic equation $x^2 + 2x - y^2 = 0$ using the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a = 1$, $b = 2$, and $c = -y^2$. Substitute these values into the formula:
$$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-y^2)}}{2 \cdot 1}$$ $$x = \frac{-2 \pm \sqrt{4 + 4y^2}}{2}$$ $$x = \frac{-2 \pm \sqrt{4(1 + y^2)}}{2}$$ $$x = \frac{-2 \pm 2\sqrt{1 + y^2}}{2}$$ $$x = -1 \pm \sqrt{1 + y^2}$$ Since $x > 0$, we choose the positive root:
$$x = -1 + \sqrt{1 + y^2}$$

Verify the inverse function by checking if $f(f^{-1}(x)) = x$. Substitute $f^{-1}(x) = -1 + \sqrt{1 + x^2}$ into $f(x)$:
$$f(f^{-1}(x)) = \sqrt{((-1 + \sqrt{1 + x^2})^2 + 2(-1 + \sqrt{1 + x^2}))}$$ Simplify the expression:
$$= \sqrt{(1 - 2\sqrt{1 + x^2} + (1 + x^2) + 2(-1 + \sqrt{1 + x^2}))}$$ $$= \sqrt{(1 + x^2)}$$ $$= x$$ Thus, the inverse function is verified.
The formula for the inverse function is $f^{-1}(x) = -1 + \sqrt{1 + x^2}$.