The function f(x) = frac{ln(pi + x)}{ln(e + x | vedclass

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

Question

(A) Let $f(x) = \frac{\ln(\pi + x)}{\ln(e + x)}$.To determine the monotonicity,we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{\

Vedclass · Accepted Answer

increasing on $[0, \infty)$

Vedclass · Answer

(A) Let $f(x) = \frac{\ln(\pi + x)}{\ln(e + x)}$.To determine the monotonicity,we find the derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{\frac{1}{\pi + x} \cdot \ln(e + x) - \frac{1}{e + x} \cdot \ln(\pi + x)}{(\ln(e + x))^2}$.The sign of $f'(x)$ depends on the numerator $g(x) = \frac{\ln(e + x)}{\pi + x} - \frac{\ln(\pi + x)}{e + x}$.Consider the function $h(t) = \frac{\ln(t)}{t}$. Its derivative is $h'(t) = \frac{1 - \ln(t)}{t^2}$.$h'(t) > 0$ for $t  e$.Since $\pi > e$,the function $h(t)$ is decreasing for $t \ge e$.Comparing $h(e+x)$ and $h(\pi+x)$,since $\pi+x > e+x > e$,we have $h(\pi+x) Thus,$\frac{\ln(\pi + x)}{\pi + x}  0$.Therefore,the function is increasing on $[0, \infty)$.

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

Similar Questions

For what values of $x$ is the function $f(x) = [x(x - 3)]^2$ an increasing function?

The function $f(x) = \log(1+x) - \frac{2x}{2+x}$ is increasing on

If $x > 0$,then $\frac{x}{1+x} - \log(1+x)$

In the open interval $\left(0, \frac{\pi}{2}\right)$, which of the following is true for the expression $\cos x + x \sin x$?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module