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

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

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

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

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

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

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

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