Science:Math Exam Resources/Courses/MATH104/December 2012/Question 01 (m)/Solution 1

To determine the values of $t$ where $f(t)$ is increasing, we must solve for the critical points of $f(t)$ and examine the sign of the derivative in between the critical points, solving $f'(t)=0$ gives

${\begin{aligned}\displaystyle f'(t)&=0\\2t\exp(t)+t^{2}\exp(t)&=0\\(t^{2}+2t)\exp(t)&=0.\end{aligned}}$

Since the exponential function is never zero for any value of $t$, we can solve the above equation by solving

${\begin{aligned}\displaystyle t^{2}+2t&=0\\t(t+2)&=0,\quad \rightarrow \quad t=0,\,-2\end{aligned}}$

Now we need to evaluate the sign of $f'(t)$ in the intervals $(-\infty ,-2),(-2,0),(0,\infty )$. Taking test points $t=-3,-1,1$ and plugging them into $f'(t)$ gives:

$\displaystyle f'(-3)=3e^{-3}>0,\quad f'(-1)=-e^{-1}<0,\quad f'(1)=3e>0.$

So $f'(t)$ is positive on $(-\infty ,-2),(0,\infty )$ and negative on $\displaystyle (-2,0)$.

Therefore, $f(t)$ is increasing for $t<-2$, $t>0$ and decreasing for $-2<t<0$.