Question

The minimum value of the function \(f(x) = x + \frac{9}{x}\), for \(x \gt 0\), is _______

• A. 0
• B. 3
• C. 6
• D. 9

Hint:

For functions of the form \(f(x) = ax + \frac{b}{x}\) with \(a, b, x \gt 0\), the minimum value is always given by \(2\sqrt{ab}\).
Applying this directly: \(2\sqrt{1 \cdot 9} = 2 \cdot 3 = 6\), which takes less than 5 seconds.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the minimum value of the single-variable function \(f(x) = x + \frac{9}{x}\) over the domain of positive real numbers, \(x \gt 0\).

Step 2: Key Formula or Approach:
There are two standard mathematical methods to solve this optimization problem:
Method 1: Calculus (Finding the critical points where the first derivative is zero and confirming the minimum using the second derivative test).
Method 2: Arithmetic Mean-Geometric Mean (AM-GM) Inequality, which states that for any non-negative real numbers \(a\) and \(b\):
\[ \frac{a+b}{2} \ge \sqrt{ab} \] with equality holding if and only if \(a = b\).

Step 3: Detailed Explanation:
Let us solve the problem using both methods to ensure robust results:

Method 1: AM-GM Inequality (Fastest)
Since \(x \gt 0\), both \(x\) and \(\frac{9}{x}\) are positive real numbers.
Applying the AM-GM inequality with \(a = x\) and \(b = \frac{9}{x}\):
\[ \frac{x + \frac{9}{x}}{2} \ge \sqrt{x \cdot \frac{9}{x}} \] The \(x\) terms inside the square root cancel out, leaving:
\[ \frac{x + \frac{9}{x}}{2} \ge \sqrt{9} \] \[ \frac{x + \frac{9}{x}}{2} \ge 3 \] Multiplying both sides by 2 gives:
\[ x + \frac{9}{x} \ge 6 \] The equality (and thus the minimum value) occurs when \(a = b\):
\[ x = \frac{9}{x} \implies x^2 = 9 \implies x = 3 \quad (\text{since } x \gt 0) \] Since \(x=3\) lies within the domain \(x \gt 0\), the minimum value is exactly \(6\).

Method 2: Calculus
Differentiate \(f(x) = x + 9x^{-1}\) with respect to \(x\):
\[ f'(x) = 1 - \frac{9}{x^2} \] To find the critical points, set \(f'(x) = 0\):
\[ 1 - \frac{9}{x^2} = 0 \implies x^2 = 9 \implies x = 3 \quad (\text{since } x \gt 0) \] Now compute the second derivative to verify the nature of this critical point:
\[ f''(x) = \frac{d}{dx}\left(1 - 9x^{-2}\right) = \frac{18}{x^3} \] At \(x = 3\):
\[ f''(3) = \frac{18}{3^3} = \frac{18}{27} = \frac{2}{3} \gt 0 \] Since the second derivative is positive, \(x = 3\) is indeed a point of local and global minimum.
The minimum value is:
\[ f(3) = 3 + \frac{9}{3} = 3 + 3 = 6 \] Both methods yield the identical minimum value of \(6\).

Step 4: Final Answer
Therefore, the minimum value of the function is \(6\), which corresponds to option (C).