Find the open intervals on which the function $f ( x ) = \frac { x } { x ^ { 2 } + 36 }$ is increasing or decreasing.
A)The function is increasing on the interval $- 6

Question

Find the open intervals on which the function $f ( x ) = \frac { x } { x ^ { 2 } + 36 }$ is increasing or decreasing.
A)The function is increasing on the interval $- 6 < x < 6$ , and decreasing on the intervals $- \infty < x < - 6$ and $6 < x < \infty$ .
B)The function is increasing on the interval $- \infty < x < - 6$ , and decreasing on the intervals $- 6 < x < 6$ and $6 < x < \infty$ .
C)The function is increasing on the interval $6 < x < \infty$ , and decreasing on the intervals $- \infty < x < - 6$ and $- 6 < x < 6$ .
D)The function is decreasing on the interval $- 6 < x < 6$ , and increasing on the intervals $- \infty < x < - 6$ and $6 < x < \infty$ .
E)The function is decreasing on the interval $- \infty < x < - 6$ , and increasing on the intervals $- 6 < x < 6$ and $6 < x < \infty$ .

Quizplus · Accepted Answer

To determine where the function is increasing or decreasing, we find its derivative and analyze its sign. The derivative of $f(x) = \frac{x}{x^2 + 36}$ using the quotient rule is $f'(x) = \frac{(x^2 + 36)(1) - (x)(2x)}{(x^2 + 36)^2} = \frac{-x^2 + 36}{(x^2 + 36)^2}$. The critical points, where $f'(x) = 0$, are at $x = \pm 6$. Testing intervals around these points, we find that $f'(x) > 0$ (indicating increasing) for $-6 < x < 6$ and $f'(x) < 0$ (indicating decreasing) for $x < -6$ and $x > 6$.

Find the Open Intervals on Which the Function f(x)=xx2+36f ( x ) = \frac { x } { x ^ { 2 } + 36 }f(x)=x2+36x​

Find the Open Intervals on Which the Function $f ( x ) = \frac { x } { x ^ { 2 } + 36 }$