Find the intervals where $f ( x ) = \frac { \log 9 x } { x }$ is increasing and where it is decreasing.
A) increasing: $( 9 e , \infty )$, decreasing: $( 0,9 e )$
B) increasing: $\left( 0 , \frac { e } { 9 } \right)$, decreasing: $\left( \frac { e } { 9 } , \infty \right)$
C) increasing: $( 90 , \infty )$, decreasing: $( 0,90 )$
D) increasing: $\left( 0 , \frac { 9 } { 10 } \right)$, decreasing: $\left( \frac { 9 } { 10 } , \infty \right)$

Question

Quizplus · Accepted Answer

To determine where the function $f(x) = \frac{\log 9x}{x}$ is increasing or decreasing, we find its derivative and analyze its sign. The derivative, using the quotient rule and the chain rule, is $f'(x) = \frac{1/x}{x} - \frac{\log 9x}{x^2} = \frac{1 - \log 9x}{x^2}$. Setting $f'(x) > 0$ to find where the function is increasing, we solve $1 - \log 9x > 0$, which simplifies to $\log 9x < 1$. Converting to exponential form gives $9x < e$, so $x < \frac{e}{9}$. Thus, the function is increasing on $\left(0, \frac{e}{9}\right)$ and decreasing on $\left(\frac{e}{9}, \infty\right)$.

Find the Intervals Where f(x)=log⁡9xxf ( x ) = \frac { \log 9 x } { x }f(x)=xlog9x​

Find the Intervals Where $f ( x ) = \frac { \log 9 x } { x }$