The least value of `alpha in R` for which `4alphax^(2)+(1)/(x)ge1,` for all `xgt0,` is

Options

A. `(1)/(64)`
B. `(1)/(32)`
C. `(1)/(27)`
D. `(1)/(25)`

Answer

To find the least value of \( \alpha \in \mathbb{R} \) for which the inequality \( 4\alpha x^2 + \frac{1}{x} \geq 1 \) holds for all \( x > 0 \), we can follow these steps:

### Step 1: Define the function
Let \( f(x) = 4\alpha x^2 + \frac{1}{x} \).

### Step 2: Find the derivative
To analyze the behavior of the function, we need to find its derivative:
\[ f'(x) = \frac{d}{dx}(4\alpha x^2) + \frac{d}{dx}\left(\frac{1}{x}\right) = 8\alpha x - \frac{1}{x^2}. \]

### Step 3: Set the derivative to zero
To find the critical points, set the derivative equal to zero:
\[ 8\alpha x - \frac{1}{x^2} = 0. \]
Rearranging gives:
\[ 8\alpha x = \frac{1}{x^2}. \]
Multiplying both sides by \( x^2 \) (since \( x > 0 \)):
\[ 8\alpha x^3 = 1. \]
Thus, we find:
\[ x^3 = \frac{1}{8\alpha} \quad \Rightarrow \quad x = \left(\frac{1}{8\alpha}\right)^{\frac{1}{3}}. \]

### Step 4: Substitute \( x \) back into the function
Now, substitute \( x \) back into \( f(x) \):
\[ f\left(\left(\frac{1}{8\alpha}\right)^{\frac{1}{3}}\right) = 4\alpha \left(\frac{1}{8\alpha}\right)^{\frac{2}{3}} + \frac{1}{\left(\frac{1}{8\alpha}\right)^{\frac{1}{3}}}. \]

Calculating each term:
1. The first term:
\[ 4\alpha \left(\frac{1}{8\alpha}\right)^{\frac{2}{3}} = 4\alpha \cdot \frac{1}{(8\alpha)^{\frac{2}{3}}} = \frac{4\alpha}{8^{\frac{2}{3}} \alpha^{\frac{2}{3}}} = \frac{4\alpha^{\frac{1}{3}}}{4} = \alpha^{\frac{1}{3}}. \]
2. The second term:
\[ \frac{1}{\left(\frac{1}{8\alpha}\right)^{\frac{1}{3}}} = 8^{\frac{1}{3}} \alpha^{\frac{1}{3}} = 2\alpha^{\frac{1}{3}}. \]

Combining these gives:
\[ f\left(\left(\frac{1}{8\alpha}\right)^{\frac{1}{3}}\right) = \alpha^{\frac{1}{3}} + 2\alpha^{\frac{1}{3}} = 3\alpha^{\frac{1}{3}}. \]

### Step 5: Set the function greater than or equal to 1
To satisfy the original inequality \( f(x) \geq 1 \) for all \( x > 0 \):
\[ 3\alpha^{\frac{1}{3}} \geq 1. \]
Dividing both sides by 3:
\[ \alpha^{\frac{1}{3}} \geq \frac{1}{3}. \]
Cubing both sides gives:
\[ \alpha \geq \left(\frac{1}{3}\right)^3 = \frac{1}{27}. \]

### Conclusion
Thus, the least value of \( \alpha \) for which the inequality holds for all \( x > 0 \) is:
\[ \alpha = \frac{1}{27}. \]