The minimum value of x^2+displaystyle1/x is:

The correct answer is Option (1) → \(\left(\displaystyle\frac{1}{2}\right)^{\frac{2}{3}}+(2)^{\frac{1}{3}}\)

Let \(f(x) = x^2 + \frac{1}{x}\). For negative x close to 0, \(\frac{1}{x}\) is a large negative number, so f has no least value over all its domain. The minimum value asked for is the local minimum, which lies at a positive x.

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

f’(x) = 0 gives

    \( \displaystyle 2x^3 = 1,\quad x = \left(\frac{1}{2}\right)^{1/3} \)

By the second derivative test:

    \( \displaystyle f''(x) = 2 + \frac{2}{x^3} = 2 + 4 = 6 \gt 0 \)

so this point is a minimum. The value there is

    \( \displaystyle f = \left(\frac{1}{2}\right)^{2/3} + \frac{1}{(1/2)^{1/3}} \)

    \( \displaystyle = \left(\frac{1}{2}\right)^{2/3} + 2^{1/3} \)