Question

The minimum value of \(x^2 + \frac{1}{x}\) is:

• A. \((4)^{\frac{2}{3}} + 2\)
• B. \(6 + (2)^{\frac{1}{3}}\)
• C. \(\left(\frac{1}{2}\right)^{\frac{1}{3}} + 5\)
• D. \(\left(\frac{1}{2}\right)^{\frac{2}{3}} + (2)^{\frac{1}{3}}\)

Answer 1

The Correct Option is D

Solution: We are tasked with finding the minimum value of the function:

\[f(x) = x^2 + \frac{1}{x}, \quad x > 0\]

Differentiating \(f(x)\): To find the critical points, compute the derivative of \(f(x)\):

\[f'(x) = 2x - \frac{1}{x^2}\]

Set \(f'(x) = 0\):

\[2x = \frac{1}{x^2}\]

Multiply through by \(x^2\) (since \(x > 0\)):

\[2x^3 = 1 \implies x^3 = \frac{1}{2} \implies x = (\frac{1}{2})^{\frac{1}{3}}\]

Computing \(f(x)\) at \(x = (\frac{1}{2})^{\frac{1}{3}}\): Substitute \(x = (\frac{1}{2})^{\frac{1}{3}}\) into \(f(x)\):

\[f\left((\frac{1}{2})^{\frac{1}{3}}\right) = \left((\frac{1}{2})^{\frac{1}{3}}\right)^2 + \frac{1}{(\frac{1}{2})^{\frac{1}{3}}}\]

Simplify each term:

1. The first term is:
2. The second term is:

Thus, the minimum value is:

\[f(x) = (\frac{1}{2})^{\frac{2}{3}} + (2)^{\frac{1}{3}}\]

Verifying it is a minimum: The second derivative of \(f(x)\) is:

\[f''(x) = 2 + \frac{2}{x^3}\]

Since \(f''(x) > 0\) for all \(x > 0\), \(f(x)\) is convex, and the critical point corresponds to a minimum.

Thus, the minimum value is:

\[\left(\frac{1}{2}\right)^{\frac{2}{3}} + (2)^{\frac{1}{3}}\]