The maximum value of $\frac{1}{x^{x}}$ is

Q: The maximum value of $\frac{1}{x^{x}}$ is

To determine the maximum value of the function $ f(x) = \frac{1}{x^x} $, we first rewrite it as $ f(x) = x^{-x} $. To find the maximum value, we need to find the critical points by taking the derivative and setting it to zero. First, take the natural logarithm of both sides to simplify differentiation: \[ \ln(f(x)) = \ln(x^{-x}) = -x \ln(x) \] Differentiate with respect to $ x $: \[ \frac{d}{dx}(-x \ln(x)) = -(\ln(x) + 1) \] Set the derivative equal to zero to find critical points: \[ -\ln(x) - 1 = 0 \quad \Rightarrow \quad \ln(x) = -1 \quad \Rightarrow \quad x = e^{-1} = \frac{1}{e} \] Now, evaluate $ f(x) $ at this critical point: \[ f\left(\frac{1}{e}\right) = \left(\frac{1}{e}\right)^{-\frac{1}{e}} = e^{\frac{1}{e}} \] Thus, the maximum value of $ \frac{1}{x^x} $ is $ e^{\frac{1}{e}} $, which corresponds to Option 4. Now, let's explain why the other options are incorrect: 1. **Option 1: e** - This option suggests that the maximum value is $ e $. However, as shown in the calculation, the maximum value is $ e^{\frac{1}{e}} $, not $ e $. 2. **Option 2: $ e^{-e} $** - This option implies a much smaller value than the actual maximum. The expression $ e^{-e} $ does not correspond to the maximum value of $ \frac{1}{x^x} $. 3. **Option 3: $ e^{-\frac{1}{e}} $** - This option suggests a value that is less than 1, whereas the actual maximum value $ e^{\frac{1}{e}} $ is greater than 1. Therefore, the correct answer is indeed Option 4: $ e^{\frac{1}{e}} $, as it accurately represents the maximum value of the function $ \frac{1}{x^x} $.

Question

Accepted Answer

To determine the maximum value of the function $ f(x) = \frac{1}{x^x} $, we first rewrite it as $ f(x) = x^{-x} $. To find the maximum value, we need to find the critical points by taking the derivative and setting it to zero.

First, take the natural logarithm of both sides to simplify differentiation:

$$
\ln(f(x)) = \ln(x^{-x}) = -x \ln(x)
$$

Differentiate with respect to $ x $:

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

Set the derivative equal to zero to find critical points:

$$
-\ln(x) - 1 = 0 \quad \Rightarrow \quad \ln(x) = -1 \quad \Rightarrow \quad x = e^{-1} = \frac{1}{e}
$$

Now, evaluate $ f(x) $ at this critical point:

$$
f\left(\frac{1}{e}\right) = \left(\frac{1}{e}\right)^{-\frac{1}{e}} = e^{\frac{1}{e}}
$$

Thus, the maximum value of $ \frac{1}{x^x} $ is $ e^{\frac{1}{e}} $, which corresponds to Option 4.

Now, let's explain why the other options are incorrect:

1. **Option 1: e** - This option suggests that the maximum value is $ e $. However, as shown in the calculation, the maximum value is $ e^{\frac{1}{e}} $, not $ e $.

2. **Option 2: $ e^{-e} $** - This option implies a much smaller value than the actual maximum. The expression $ e^{-e} $ does not correspond to the maximum value of $ \frac{1}{x^x} $.

3. **Option 3: $ e^{-\frac{1}{e}} $** - This option suggests a value that is less than 1, whereas the actual maximum value $ e^{\frac{1}{e}} $ is greater than 1.

Therefore, the correct answer is indeed Option 4: $ e^{\frac{1}{e}} $, as it accurately represents the maximum value of the function $ \frac{1}{x^x} $.