Science:Math Exam Resources/Courses/MATH100/December 2015/Question 10 (g)/Solution 1

We need to determine the sign of${\displaystyle f'(x)=-e^{-x}\left(x^{2}+x-1\right)}$.

From part (e) we have ${\displaystyle f'(x)=-e^{-x}\left(x^{2}+x-1\right)}$, where ${\displaystyle e^{-x}}$ is always positive so it doesn't have any role in the sign. We only need to determine the sign of ${\displaystyle -x^{2}-x+1}$.

From part (f) ${\displaystyle -x^{2}-x+1}$ is ${\displaystyle 0\ \quad }$ at ${\displaystyle x={\dfrac {-1+{\sqrt {5}}}{2}}\approx 0.35,\ {\text{and}}\ \ x={\dfrac {-1-{\sqrt {5}}}{2}}\approx -0.13}$,

so we make a sign chart for ${\displaystyle f'(x)}$ to find where it is positive and negative and therefore where the function ${\displaystyle f(x)}$ is increasing and decreasing.

 ${\displaystyle \displaystyle (-\infty ,{\dfrac {-1-{\sqrt {5}}}{2}})}$ ${\displaystyle \displaystyle ({\dfrac {-1-{\sqrt {5}}}{2}},{\dfrac {-1+{\sqrt {5}}}{2}})}$ ${\displaystyle \displaystyle ({\dfrac {-1+{\sqrt {5}}}{2}},\infty )}$ ${\displaystyle -x^{2}-x+1}$ ${\displaystyle \displaystyle -}$ ${\displaystyle \displaystyle +}$ ${\displaystyle \displaystyle -}$ ${\displaystyle f'(x)=e^{-x}\left(-x^{2}-x+1\right)}$ ${\displaystyle \displaystyle -}$ ${\displaystyle \displaystyle +}$ ${\displaystyle \displaystyle -}$ ${\displaystyle f(x)}$ decreasing increasing decreasing

Intervals that ${\displaystyle f(x)}$ is decreasing: ${\displaystyle {\color {blue}(-\infty ,{\dfrac {-1-{\sqrt {5}}}{2}}),\ ({\dfrac {-1+{\sqrt {5}}}{2}},\infty )}}$.

Intervals that ${\displaystyle f(x)}$ is increasing: ${\displaystyle {\color {blue}({\dfrac {-1-{\sqrt {5}}}{2}},{\dfrac {-1+{\sqrt {5}}}{2}})}}$.