To find the x-coordinate of the point on the parabola $f(x) = x^2 - 1$ that is closest to the origin, we need to minimize the distance from the point on the curve to the origin. The distance $d$ from a point $(x, f(x))$ to the origin $(0, 0)$ is given by the distance formula:

$d(x) = \sqrt{x^2 + [f(x)]^2}$

Substitute $f(x) = x^2 - 1$ into this formula:

$d(x) = \sqrt{x^2 + (x^2 - 1)^2}$

To simplify the problem, we can minimize the square of the distance $d^2(x)$, since the square root function is monotonically increasing:

$d^2(x) = x^2 + (x^2 - 1)^2$

Expanding $(x^2 - 1)^2$:

$d^2(x) = x^2 + (x^4 - 2x^2 + 1)$

Simplifying further:

$d^2(x) = x^4 - x^2 + 1$

Now, to minimize $d^2(x)$, we take the derivative of $d^2(x)$ with respect to $x$ and set it equal to zero:

$\frac{d}{dx} \left( x^4 - x^2 + 1 \right) = 4x^3 - 2x$

Set the derivative equal to zero to find the critical points:

$4x^3 - 2x = 0$

Factor the equation:

$2x(2x^2 - 1) = 0$

This gives two solutions:

$x = 0 \quad \text{or} \quad 2x^2 - 1 = 0$

Solving $2x^2 - 1 = 0$:

$x^2 = \frac{1}{2} \quad \Rightarrow \quad x = \pm \frac{\sqrt{2}}{2}$

Thus, the critical points are $x = 0$, $x = \frac{\sqrt{2}}{2}$, and $x = -\frac{\sqrt{2}}{2}$.

To determine which point minimizes the distance, we calculate the second derivative of $d^2(x)$:

$\frac{d^2}{dx^2} \left( x^4 - x^2 + 1 \right) = 12x^2 - 2$

Evaluate the second derivative at the critical points:

• For $x = 0$, $\frac{d^2}{dx^2} = -2$, indicating a local maximum.
• For $x = \pm \frac{\sqrt{2}}{2}$, $\frac{d^2}{dx^2} = 12 \times \left( \frac{1}{2} \right) - 2 = 4$, indicating a local minimum.

Therefore, the point on the parabola closest to the origin is at $x = 0$.

Conclusion: The x-coordinate of the point on the parabola $f(x) = x^2 - 1$ that is closest to the origin is $\boxed{0}$.

Would you like further clarification or details on this process?

Here are 5 related questions:

1. How do you find the distance from a point to a curve in general?
2. What happens if the second derivative is negative at a critical point?
3. How do you interpret the first and second derivatives in terms of graph behavior?
4. Can this method be applied to other types of curves, such as a circle or a line?
5. How do you find the closest point on a curve to a point not on the curve?

Tip: Always start by minimizing the square of the distance, as it simplifies calculations and avoids dealing with square roots.