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 $f^{\prime }(t)=0$ gives

$\begin{array}{rl}{f}^{\prime }(t)& =0\\ 2t\exp (t)+t^2\exp (t)& =0\\ (t^2+2t)\exp (t)& =0.\end{array}$

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

$\begin{array}{rl}{t}^2+2t& =0\\ t(t+2)& =0,\to t=0,-2\end{array}$

Now we need to evaluate the sign of $f^{\prime }(t)$ in the intervals $(-\mathrm{\infty },-2),(-2,0),(0,\mathrm{\infty })$. Taking test points $t=-3,-1,1$ and plugging them into $f^{\prime }(t)$ gives:

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

So $f^{\prime }(t)$ is positive on $(-\mathrm{\infty },-2),(0,\mathrm{\infty })$ and negative on $(-2,0)$.

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