The set of real values of x for which f(x) = | vedclass

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

Question

(A) Given the function $f(x) = \frac{x}{\log x}$.First,we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{(\log x)(1) - (x)(\frac{1

Vedclass · Accepted Answer

$\{x: x \geq e\}$

Vedclass · Answer

(A) Given the function $f(x) = \frac{x}{\log x}$.First,we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{(\log x)(1) - (x)(\frac{1}{x})}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2}$.For the function to be increasing,we must have $f'(x) > 0$.Since $(\log x)^2$ is always positive for $x > 0$ and $x 
eq 1$,the condition $f'(x) > 0$ implies $\log x - 1 > 0$.$\log x > 1$.Since the base of the logarithm is $e$,we have $\log_e x > \log_e e$.Therefore,$x > e$.Thus,the set of values is $\{x: x > e\}$.

The set of real values of $x$ for which $f(x) = \frac{x}{\log x}$ is increasing,is

Similar Questions

The curve represented by $x = t^5 + 5t^3 + 20t + 7$ and $y = 4t^3 - 3t^2 - 18t + 3$ is decreasing in the interval

The function $f(x) = \frac{\ln(\pi+x)}{\ln(e+x)}$ is

If $\log (1+x)-\frac{2x}{2+x}$ is increasing,then

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

The function $f(x) = x + \frac{1}{x}, (x \neq 0)$ is a non-increasing function in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module