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

MATH104 December 2012

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Question 01 (m)
On what intervals is the function $\displaystyle {}f(t)=t^{2}\exp(t)$ increasing? Decreasing?

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Hint
To determine the interval(s) where a function $f(t)$ is increasing, we must search for the values of $t$ where the derivative of $f(t)$ has what quality? What about for intervals where the function is decreasing?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $-2<t<0$.

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