Consider the fucntion `f(x)=((1)/(x))^(2x^2)` , where `xgt0`. At what value of x does the function attain maximum value ?

Options

A. e
B. `sqrt(e )`
C. `(1)/sqrt(e )`
D. `(1)/(e )`

Explanation

To find the value of \( x \) at which the function \( f(x) = \left( \frac{1}{x} \right)^{2x^2} \) attains its maximum value, we will follow these steps:

### Step 1: Rewrite the function using logarithms
Let \( y = f(x) = \left( \frac{1}{x} \right)^{2x^2} \). Taking the natural logarithm of both sides, we have:
\[ \log y = 2x^2 \log \left( \frac{1}{x} \right) \]
Using the property of logarithms, \( \log \left( \frac{1}{x} \right) = -\log x \), we can rewrite this as:
\[ \log y = -2x^2 \log x \]

### Step 2: Differentiate the logarithmic form
Now, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\log y) = \frac{1}{y} \frac{dy}{dx} \]
Using the product rule on the right side:
\[ \frac{d}{dx}(-2x^2 \log x) = -2 \left( 2x \log x + x^2 \cdot \frac{1}{x} \right) = -2(2x \log x + x) = -4x \log x - 2x \]
Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = -4x \log x - 2x \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \left( -4x \log x - 2x \right) \]
Substituting back \( y = \left( \frac{1}{x} \right)^{2x^2} \):
\[ \frac{dy}{dx} = \left( \frac{1}{x} \right)^{2x^2} \left( -4x \log x - 2x \right) \]

### Step 4: Set the derivative to zero
To find the critical points, we set \( \frac{dy}{dx} = 0 \):
\[ -4x \log x - 2x = 0 \]
Factoring out \( -2x \):
\[ -2x(2 \log x + 1) = 0 \]
Since \( x > 0 \), we can ignore \( -2x = 0 \), leading to:
\[ 2 \log x + 1 = 0 \]
Thus,
\[ \log x = -\frac{1}{2} \]
Exponentiating both sides gives:
\[ x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \]

### Step 5: Determine if it is a maximum
To confirm that this point is a maximum, we can analyze the sign of \( \frac{dy}{dx} \) around \( x = \frac{1}{\sqrt{e}} \):
- For \( x < \frac{1}{\sqrt{e}} \), \( \log x > -\frac{1}{2} \) implies \( 2 \log x + 1 > 0 \), hence \( \frac{dy}{dx} > 0 \) (function is increasing).
- For \( x > \frac{1}{\sqrt{e}} \), \( \log x < -\frac{1}{2} \) implies \( 2 \log x + 1 < 0 \), hence \( \frac{dy}{dx} < 0 \) (function is decreasing).

Since the function increases before \( x = \frac{1}{\sqrt{e}} \) and decreases after, we conclude that \( x = \frac{1}{\sqrt{e}} \) is indeed a maximum.

### Final Answer
The function \( f(x) \) attains its maximum value at:
\[ x = \frac{1}{\sqrt{e}} \]