Questions: Given the graph of y=4=x^2 defined on 0 ≤ x ≤ 2 (See graph). The distance from the point (0,2) to a point on the graph is given by the equation: d(x)=sqrt(x^4-3 x^2+4), Find the x value of the point on the graph that would be closest to (0,2)

Solution Steps

We are given the distance function $d(x) = \sqrt{x^4 - 3x^2 + 4}$. To minimize the distance, we can equivalently minimize the square of the distance, which simplifies calculations. Let $D(x) = d(x)^2 = x^4 - 3x^2 + 4$. Then $D'(x) = 4x^3 - 6x$.

To find the critical points, we set the derivative equal to zero: $4x^3 - 6x = 0$. Factoring gives $2x(2x^2 - 3) = 0$. Thus, $x=0$ or $2x^2 - 3 = 0$, implying $x^2 = \frac{3}{2}$, so $x = \pm\sqrt{\frac{3}{2}}$. Since the problem specifies $0 \le x \le 2$, we consider only $x = 0$ and $x = \sqrt{\frac{3}{2}}$.

We evaluate the distance function at the critical points and endpoints: $d(0) = \sqrt{0 - 0 + 4} = 2$ $d(\sqrt{\frac{3}{2}}) = \sqrt{(\frac{3}{2})^2 - 3(\frac{3}{2}) + 4} = \sqrt{\frac{9}{4} - \frac{9}{2} + 4} = \sqrt{\frac{9 - 18 + 16}{4}} = \sqrt{\frac{7}{4}} = \frac{\sqrt{7}}{2} \approx 1.32$ $d(2) = \sqrt{16 - 12 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.83$

The minimum distance occurs at $x = \sqrt{\frac{3}{2}}$.

Final Answer