The interval in which the function f(x) = x^x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the function $f(x) = x^x$ for $x > 0$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = x \ln(x)$.Differentiating both sides wi

Vedclass · Accepted Answer

$\left[\frac{1}{e}, \infty\right)$

Vedclass · Answer

(D) Given the function $f(x) = x^x$ for $x > 0$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = x \ln(x)$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{f(x)} f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$.Thus,$f'(x) = x^x(1 + \ln(x))$.For the function to be strictly increasing,we require $f'(x) > 0$.Since $x^x > 0$ for all $x > 0$,the condition $f'(x) > 0$ implies $1 + \ln(x) > 0$.$\ln(x) > -1$.$x > e^{-1}$,which means $x > \frac{1}{e}$.Therefore,the interval in which the function is strictly increasing is $\left(\frac{1}{e}, \infty\right)$.Note: The option $\left[\frac{1}{e}, \infty\right)$ is the standard representation for the interval where the function is non-decreasing.

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

Similar Questions

Let $f(x) = \int \frac{x^2-3x+2}{x^4+1} \, dx$. Then the function decreases in the interval:

The function $f(x) = \tan^{-1}(\sin x + \cos x)$ is an increasing function in the interval.....

If $f(x) = \frac{x}{x^2+1}$ is an increasing function,then the value of $x$ lies in:

Let $g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)$ and $f^{\prime \prime}(x)>0$ for all $x \in(0,3)$. If $g$ is decreasing in $(0, \alpha)$ and increasing in $(\alpha, 3)$,then $8 \alpha$ is

The function $f$ defined by $f(x) = (x + 2) e^{-x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module