Question

If \( f(x) = x^x \), then the interval in which \( f(x) \) decreases is:

• A. \( \left[ 0, \frac{1}{e} \right] \)
• B. \( \left[ 0, e \right] \)
• C. \( \left[ \frac{1}{e}, \infty \right] \)
• D. \( \left[ 0, e^e \right] \)

Hint:

To find increasing or decreasing intervals, differentiate the function and analyze where \( f'(x) \) is positive or negative.

Answer 1

The Correct Option is A

We are given the function: \[ f(x) = x^x. \] Step 1: Differentiating \( f(x) = x^x \)
Taking the natural logarithm on both sides: \[ y = x^x \Rightarrow \ln y = x \ln x. \] Differentiating both sides using implicit differentiation: \[ \frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{x} + \ln x \cdot 1. \] \[ \frac{dy}{dx} = x^x \left( 1 + \ln x \right). \] Step 2: Finding the critical points
Setting \( \frac{dy}{dx} = 0 \): \[ x^x \left( 1 + \ln x \right) = 0. \] Since \( x^x \neq 0 \) for \( x > 0 \), we solve: \[ 1 + \ln x = 0. \] \[ \ln x = -1. \] \[ x = \frac{1}{e}. \] Step 3: Finding the decreasing interval
- For \( x < \frac{1}{e} \), we check the sign of \( \frac{dy}{dx} \): \[ 1 + \ln x < 0 \Rightarrow \text{Negative derivative} \Rightarrow \text{Decreasing}. \] - For \( x > \frac{1}{e} \): \[ 1 + \ln x > 0 \Rightarrow \text{Positive derivative} \Rightarrow \text{Increasing}. \] Thus, \( f(x) \) decreases in: \[ \left[ 0, \frac{1}{e} \right]. \]

Answer 2

To determine the interval where the function \( f(x) = x^x \) decreases, we first find the derivative, since a function decreases when its derivative is less than zero.

Rewrite the function as \( f(x) = e^{x \ln x} \). Using the natural logarithm:

\( \ln y = x \ln x \)

Differentiating both sides with respect to \( x \) using implicit differentiation:

\( \frac{dy}{dx} = \frac{d}{dx}(e^{x \ln x}) = e^{x \ln x} \cdot \frac{d}{dx}(x \ln x) \)

Now, differentiate \( x \ln x \):

\( \frac{d}{dx}(x \ln x) = \ln x + 1 \)

Thus, the derivative becomes:

\( \frac{dy}{dx} = x^x (\ln x + 1) \)

To find where \( f(x) \) decreases, solve \( \frac{dy}{dx} < 0 \):

\( x^x (\ln x + 1) < 0 \)

Since \( x^x > 0 \) for all \( x > 0 \), the sign of the derivative depends on \( \ln x + 1 \):

Solve \( \ln x + 1 < 0 \Rightarrow \ln x < -1 \Rightarrow x < \frac{1}{e} \)

However, the domain of \( f(x) = x^x \) is \( x > 0 \), so the decreasing interval is:

\[ \boxed{\left[0, \frac{1}{e}\right]} \]