The least value of the function `f(x) = ax + (b)/(x) (x gt 0, a gt 0, b gt 0)`

Options

A. `sqrt(ab)`
B. `2 sqrt(ab)`
C. ab
D. 2ab

Answer

To find the least value of the function \( f(x) = ax + \frac{b}{x} \) where \( x > 0 \), \( a > 0 \), and \( b > 0 \), we can follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(ax + \frac{b}{x}) = a - \frac{b}{x^2} \]

### Step 2: Set the derivative to zero
To find the critical points, we set the first derivative equal to zero:

\[ f'(x) = 0 \implies a - \frac{b}{x^2} = 0 \]

Rearranging gives:

\[ \frac{b}{x^2} = a \implies x^2 = \frac{b}{a} \]

Taking the square root (considering only the positive root since \( x > 0 \)):

\[ x = \sqrt{\frac{b}{a}} \]

### Step 3: Find the second derivative
Next, we need to determine whether this critical point is a minimum or maximum by finding the second derivative:

\[ f''(x) = \frac{d}{dx}(a - \frac{b}{x^2}) = \frac{2b}{x^3} \]

### Step 4: Evaluate the second derivative at the critical point
Now we evaluate \( f''(x) \) at \( x = \sqrt{\frac{b}{a}} \):

\[ f''\left(\sqrt{\frac{b}{a}}\right) = \frac{2b}{\left(\sqrt{\frac{b}{a}}\right)^3} = \frac{2b}{\frac{b^{3/2}}{a^{3/2}}} = \frac{2a^{3/2}}{b^{1/2}} \]

Since \( a > 0 \) and \( b > 0 \), \( f''\left(\sqrt{\frac{b}{a}}\right) > 0 \), indicating that we have a local minimum.

### Step 5: Calculate the minimum value
Finally, we substitute \( x = \sqrt{\frac{b}{a}} \) back into the original function to find the minimum value:

\[ f\left(\sqrt{\frac{b}{a}}\right) = a\left(\sqrt{\frac{b}{a}}\right) + \frac{b}{\sqrt{\frac{b}{a}}} \]

This simplifies to:

\[ = a \cdot \sqrt{\frac{b}{a}} + b \cdot \sqrt{\frac{a}{b}} = \sqrt{ab} + \sqrt{ab} = 2\sqrt{ab} \]

### Conclusion
Thus, the least value of the function \( f(x) = ax + \frac{b}{x} \) is:

\[ \boxed{2\sqrt{ab}} \]