In which interval is the function f(x) = x^x | vedclass

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

Question

(C) Let $y = x^x$. Taking the logarithm on both sides,we get $\log y = x \log x$. Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{d

Vedclass · Accepted Answer

$(0, 1/e)$

Vedclass · Answer

(C) Let $y = x^x$. Taking the logarithm on both sides,we get $\log y = x \log x$. Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{dy}{dx} = x^x(1 + \log x)$. For the function to be decreasing,we must have $\frac{dy}{dx} Since $x^x > 0$ for $x > 0$,the condition $\frac{dy}{dx} This simplifies to $\log x Therefore,$x Since the domain of $x^x$ is $x > 0$,the interval in which the function is decreasing is $(0, 1/e)$.

In which interval is the function $f(x) = x^x$ a decreasing function?

Similar Questions

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

If $y = 2x + \cot^{-1} x + \log(\sqrt{1 + x^2} - x)$,then $y$

For what values of $x$ does the function $f(x) = (x - 1)^2 (x - 2)$ monotonically decrease?

For every value of $x$,the function $f(x)=\frac{1}{a^{x}}, a>0$ is

For every value of $x$,the function $f(x) = e^x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module