The interval in which the function f(x) = x^x, x 0, is strictly increasing is

Q: Wait, why can't I just use the power rule $nx^{n-1}$?

The power rule is only for functions where the exponent is a constant. Here, the exponent is a variable $x$. You must use logarithmic differentiation to handle the variable exponent.

Q: Why is $x^x$ always positive for $x > 0$?

For any positive base $x$, raising it to any real power $x$ will always yield a positive result. Think of it as $e^{x \ln x}$. Since the exponential function $e^u$ is always positive for any real $u$, $x^x$ must be positive.

Q: Does the interval include $1/e$?

Yes, the function is strictly increasing on the interval $[1/e, \infty)$. At $x = 1/e$, the derivative is zero, which satisfies the condition for a function to be non-decreasing, and since it is strictly increasing for all $x > 1/e$, the closed interval is correct.

Question

The interval in which the function $f(x) = x^x, x > 0$, is strictly increasing is

Options:

(A) $(0, \frac{1}{e}]$

(B) $(0, \infty)$

(C) $[\frac{1}{e}, \infty)$

(D) $[\frac{1}{e^2}, 1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before the graph of $f(x) = x^x$. It is a fascinating curve—it dips down, reaches a minimum, and then climbs steadily toward infinity.

When you first encounter $f(x) = x^x$, your intuition might scream "power rule!" or "exponential rule!". However, the power rule $nx^{n-1}$ only works when the exponent is a constant, and the exponential rule $a^x \ln a$ only works when the base is a constant.

Here, both are variables. This is a hybrid beast. To tame it, we need a special tool: **logarithmic differentiation**. By taking the natural logarithm of both sides, we transform the exponent into a coefficient, turning a complex power into a simple product.

### The Calculus Battle

Once we have $\ln(f(x)) = x \ln x$, the path forward becomes clear. We differentiate both sides with respect to $x$.

On the left, we use the chain rule to get:

$$\frac{f'(x)}{f(x)}$$

On the right, we apply the product rule: the derivative of $x$ is $1$, and the derivative of $\ln x$ is $\frac{1}{x}$. This gives us $\ln x + x(\frac{1}{x})$, which simplifies beautifully to $\ln x + 1$.

Now, we isolate $f'(x)$ by multiplying by $f(x)$, which is just our original function $x^x$. Thus, we arrive at the elegant derivative:

$$f'(x) = x^x(1 + \ln x)$$

### The Critical Point

Now, we return to our goal: finding where the function is strictly increasing. This happens when the slope is non-negative, or $f'(x) \ge 0$.

Look closely at our derivative:

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

Because $x > 0$, the term $x^x$ is always positive—it can never drag our slope into the negative. This means the sign of the entire derivative depends solely on the term $(1 + \ln x)$.

We solve the inequality $1 + \ln x \ge 0$, which leads us to $\ln x \ge -1$. Exponentiating both sides, we find $x \ge e^{-1}$, or:

$$x \ge \frac{1}{e}$$

The function is strictly increasing for all values of $x$ in the interval $

Animated Solution for Mathematics - Differentiation: The interval in which the function $f (x) = x^{x}, x > 0$ , is strictly increasing is

Select Answer:

Visualized Solution

Defining the Function $f (x) = x^{x}$

Condition for Strictly Increasing

Logarithmic Differentiation Setup

Differentiating Both Sides

Applying the Product Rule

Simplifying the Derivative

Isolating $f^{'} (x)$

Applying the Increasing Condition

Analyzing the Factors

Solving the Inequality

Exponentiating to Find $x$

Final Interval

The Sigma Insight: Monotonicity

Analyzing the Setup

The Calculus Battle

The Critical Point

Similar Questions

Let $f (x) = {x e^{a x}, x + a x^{2} - x^{3}, x \leq 0 x > 0$ . Where $a$ is a positive constant. Find the interval in which $f^{'} (x)$ is increasing.

If $f : R \to R$ is a differentiable function such that $f^{'} (x) > 2 f (x)$ for all $x \in R$ , and $f (0) = 1$ , then

Let $f$ be a real valued function, defined on $R - {- 1, 1}$ and given by $f (x) = 3 lo g_{e} \frac{x - 1}{x + 1} - \frac{2}{x - 1}$ . Then in which of the following intervals, function $f (x)$ is increasing?

The function $f (x) = x e^{x (1 - x)}, x \in R$ , is

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

Let $f (x) = e^{x} - x$ and $g (x) = x^{2} - x, x \in R$ . Then the set of all $x \in R$ , where the function $h (x) = (f o g) (x)$ is increasing, is :

If $f (x) = x e^{x (1 - x)}$ , then $f (x)$ is

The function, $f (x) = (3 x - 7) x^{2/3}$ , $x \in R$ , is increasing for all $x$ lying in :

If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in

The function $y = 2 x^{2} - ln ∣ x ∣$ is monotonically increasing for values of $x (\neq = 0)$ satisfying the inequalities ...... and monotonically decreasing for values of $x$ satisfying the inequalities .........

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The interval in which the function f(x)=xx,x>0, is strictly increasing is