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

Answer: (D)

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

Answer 2

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:

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