Question

The maximum value of $x^{-x}$ is.

• A. $e^{-e}$
• B. $e^{1/e}$
• C. $e^{-1/e}$
• D. e

Answer 1

Answer: (B)

$e^{1/e}$

Answer 2

To find the maximum value of the function $f(x) = x^{-x}$, we use calculus, specifically differentiation.

1. Define the function:
   Let $y = x^{-x}$.
   For the function to be well-defined and differentiable using logarithms, we consider $x > 0$.

2. Apply Natural Logarithm:
   Taking the natural logarithm on both sides simplifies the exponent:
   $\ln y = \ln(x^{-x})$
   Using the logarithm property $\ln(a^b) = b \ln a$:
   $\ln y = -x \ln x$

3. Differentiate with respect to x:
   Differentiate both sides implicitly with respect to $x$. Use the product rule for the right side: $(uv)' = u'v + uv'$.
   $\frac{1}{y} \frac{dy}{dx} = -\left( \frac{d}{dx}(x) \cdot \ln x + x \cdot \frac{d}{dx}(\ln x) \right)$
   $\frac{1}{y} \frac{dy}{dx} = -\left( 1 \cdot \ln x + x \cdot \frac{1}{x} \right)$
   $\frac{1}{y} \frac{dy}{dx} = -(\ln x + 1)$

4. Solve for $\frac{dy}{dx}$:
   Multiply both sides by $y$:
   $\frac{dy}{dx} = -y(\ln x + 1)$
   Substitute $y = x^{-x}$ back into the equation:
   $\frac{dy}{dx} = -x^{-x}(\ln x + 1)$

5. Find Critical Points:
   To find the maximum or minimum values, set the first derivative equal to zero:
   $\frac{dy}{dx} = 0$
   $-x^{-x}(\ln x + 1) = 0$
   Since $x^{-x} = \frac{1}{x^x}$ is always positive for $x > 0$, we must have:
   $\ln x + 1 = 0$
   $\ln x = -1$
   To solve for $x$, exponentiate both sides with base $e$:
   $x = e^{-1}$
   $x = \frac{1}{e}$

6. Determine if it's a Maximum:
   We use the first derivative test. We examine the sign of $\frac{dy}{dx}$ around $x = \frac{1}{e}$.
   Recall $\frac{dy}{dx} = -x^{-x}(\ln x + 1)$. The term $-x^{-x}$ is always negative. The sign of $\frac{dy}{dx}$ depends on the sign of $(\ln x + 1)$.

   • For $x < \frac{1}{e}$ (e.g., $x = \frac{1}{e^2}$):
     $\ln x < \ln(\frac{1}{e}) \implies \ln x < -1$.
     So, $\ln x + 1 < 0$.
     Then $\frac{dy}{dx} = (\text{negative}) \times (\text{negative}) = \text{positive}$.
     This means $f(x)$ is increasing for $x < \frac{1}{e}$.

   • For $x > \frac{1}{e}$ (e.g., $x = e$):
     $\ln x > \ln(e) \implies \ln x > 1$.
     So, $\ln x + 1 > 0$.
     Then $\frac{dy}{dx} = (\text{negative}) \times (\text{positive}) = \text{negative}$.
     This means $f(x)$ is decreasing for $x > \frac{1}{e}$.

Since the function changes from increasing to decreasing at $x = \frac{1}{e}$, this point corresponds to a local maximum.

7. Calculate the Maximum Value:
   Substitute $x = \frac{1}{e}$ back into the original function $f(x) = x^{-x}$:
   $f\left(\frac{1}{e}\right) = \left(\frac{1}{e}\right)^{-\frac{1}{e}}$
   Since $\frac{1}{e} = e^{-1}$, we have:
   $f\left(\frac{1}{e}\right) = (e^{-1})^{-\frac{1}{e}}$
   Using the exponent rule $(a^b)^c = a^{bc}$:
   $f\left(\frac{1}{e}\right) = e^{(-1) \cdot (-\frac{1}{e})}$
   $f\left(\frac{1}{e}\right) = e^{\frac{1}{e}}$

The maximum value of $x^{-x}$ is $e^{1/e}$.