Using the derivative of f(x) given below, determine the intervals on which f(x) is increasing or decreasing.
-$$f ^ { \prime } ( x ) = ( x - 6 ) e ^ { - x }$$
A) Decreasing on $( - \infty , - 6 )$; increasing on $( - 6 , \infty )$
B) Increasing on $( - \infty , - 6 )$; decreasing on $( - 6 , \infty )$
C) Increasing on $( - \infty , 6 )$; decreasing on $( 6 , \infty )$
D) Decreasing on $( - \infty , 6 )$; increasing on $( 6 , \infty )$

Question

Quizplus · Accepted Answer

To determine intervals where a function is increasing or decreasing, we need to analyze the sign of the derivative. In this case, $f'(x)=(x-6)e^{-x}$. To find the intervals where $f'(x)>0$ (increasing), we need to solve the inequality $(x-6)e^{-x}>0$, which is equivalent to $x-6>0$ and $e^{-x}>0$ or $x-6<0$ and $e^{-x}<0$. The latter is not possible since $e^{-x}$ is always positive, so let's focus on the former: \begin{align*} x-6 &>0 \ x &>6 \end{align*} Therefore, $f'(x)>0$ when $x>6$, which means that $f(x)$ is increasing on $(6,\infty)$. To find the intervals where $f'(x)<0$ (decreasing), we need to solve the inequality $(x-6)e^{-x}<0$, which is equivalent to $x-6<0$ and $e^{-x}>0$ or $x-6>0$ and $e^{-x}<0$. The former is not possible since $e^{-x}$ is always positive, so let's focus on the latter: \begin{align*} x-6 &>0 \ x &<6 \end{align*} Therefore, $f'(x)<0$ when $x<6$, which means that $f(x)$ is decreasing on $(-\infty,6)$. Thus, the correct choice is D: Decreasing on $(-\infty,6)$; increasing on $(6,\infty)$.

Using the Derivative of F(x) Given Below, Determine the Intervals f′(x)=(x−6)e−xf ^ { \prime } ( x ) = ( x - 6 ) e ^ { - x }f′(x)=(x−6)e−x

Using the Derivative of F(x) Given Below, Determine the Intervals $f ^ { \prime } ( x ) = ( x - 6 ) e ^ { - x }$