Question

If the expression \(x + \frac{1}{x^2},\; x>0\) attains minimum value at \(x=\alpha\), then \(\alpha^3\) is:

Hint:

For \(x + \frac{1}{x^n}\) with \(x>0\), differentiate and solve \(f'(x)=0\) to find the minimum point.

Answer 1

Correct Answer: 2

Concept: Use differentiation to find point of minimum for \(x>0\).

Step 1:Define the function: \[ f(x) = x + \frac{1}{x^2} \]

Step 2:Differentiate with respect to \(x\): \[ f'(x) = 1 - \frac{2}{x^3} \]

Step 3:Set \(f'(x) = 0\) for critical point: \[ 1 - \frac{2}{x^3} = 0 \] \[ 1 = \frac{2}{x^3} \] \[ x^3 = 2 \]

Step 4:Since \(x>0\): \[ x = 2^{1/3} = \alpha \] \[ \alpha^3 = 2 \]

Step 5:Verify minima using second derivative: \[ f''(x) = \frac{6}{x^4}>0 \text{ for } x>0 \] Hence, it is a point of minimum. \[ \Rightarrow \alpha^3 = 2 \]