2.9 Q-5

Question Statement

Show that the function $y = x^x$ has a minimum value at $x = \frac{1}{e}$.

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Background and Explanation

In this problem, we will:

1. Take the natural logarithm of both sides to simplify the function.
2. Differentiate $y = x^x$ to find the critical points where the first derivative equals zero.
3. Use the second derivative test to confirm that the critical point at $x = \frac{1}{e}$ corresponds to a minimum.

Concepts involved include:

• The natural logarithm properties, particularly $\ln e = 1$.
• The second derivative test to confirm the nature (minimum or maximum) of a critical point.

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Solution

Step 1: Given Function

We are given:

$$y = x^x$$

Step 2: Take the Natural Logarithm of Both Sides

Taking the natural logarithm of both sides:

$$\ln y = \ln(x^x) = x \ln x$$

Step 3: Differentiate Implicitly

Now, differentiate both sides with respect to $x$:

$$\frac{d}{dx} [\ln y] = \frac{d}{dx} [x \ln x]$$

Using the chain rule on the left side and product rule on the right side:

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

Solving for $\frac{dy}{dx}$:

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

Substitute $y = x^x$:

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

Step 4: Set the First Derivative Equal to Zero

To find the critical points, set $\frac{dy}{dx} = 0$:

$$x^x (\ln x + 1) = 0$$

Since $x^x \neq 0$ for any positive $x$, we have:

$$\ln x + 1 = 0$$

Solving for $x$:

$$\ln x = -1$$ $$x = e^{-1} = \frac{1}{e}$$

Step 5: Compute the Second Derivative

Now, compute the second derivative to confirm the nature of the critical point:

$$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left[ x^x (1 + \ln x) \right]$$

Using the product rule:

$$\frac{d^2 y}{dx^2} = x^x \left[ (1 + \ln x)^2 + \frac{1}{x} \right]$$

Substitute $x = \frac{1}{e}$:

$$\frac{d^2 y}{dx^2} \Bigg|_{x = \frac{1}{e}} = \left( \frac{1}{e} \right)^{\frac{1}{e}} \left[ (1 - 1)^2 + e \right]$$

Simplify:

$$\frac{d^2 y}{dx^2} \Bigg|_{x = \frac{1}{e}} = \left( \frac{1}{e} \right)^{\frac{1}{e}} \times e$$

Since $e > 0$, the second derivative is positive, confirming that $x = \frac{1}{e}$ is a minimum.

Thus, the function $y = x^x$ has its minimum value at $x = \frac{1}{e}$.

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Key Formulas or Methods Used

1. Natural Logarithm:

$$\ln(x^x) = x \ln x$$

2. First Derivative:

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

3. Second Derivative Test: If $\frac{d^2y}{dx^2} > 0$ at a critical point, the function has a minimum at that point.

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Summary of Steps

1. Start with the given function $y = x^x$.
2. Take the natural logarithm of both sides: $\ln y = x \ln x$.
3. Differentiate implicitly to find the first derivative: $\frac{dy}{dx} = x^x (\ln x + 1)$.
4. Set the first derivative equal to zero to find the critical point: $x = \frac{1}{e}$.
5. Compute the second derivative and confirm it is positive at $x = \frac{1}{e}$, confirming a minimum.