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

Step 1: Understanding the Concept:
To find the minimum value of a function, we take its first derivative and set it to zero. Alternatively, we can use the AM-GM inequality for positive terms.
Step 2: Detailed Explanation:
Let \(f(x) = x + \frac{1}{x^{2}}\) for \(x>0\).
Using Differentiation:
\[ f'(x) = 1 - \frac{2}{x^{3}} \]
For minima, set \(f'(x) = 0\):
\[ 1 - \frac{2}{x^{3}} = 0 \implies \frac{2}{x^{3}} = 1 \implies x^{3} = 2 \]
The second derivative \(f''(x) = \frac{6}{x^{4}}\) is positive for all \(x>0\), confirming that the value at \(x = \sqrt[3]{2}\) is indeed a minimum.
Given that the minimum is at \(x = \alpha\), we have \(\alpha^{3} = 2\).
Step 3: Final Answer:
The value of \(\alpha^{3}\) is 2.