describe the long run behavior of ( f(t) = -5left(\frac{1}{3}\right)^{t}+3 ):\nas ( t \to -infty, f(t) \to…

Explanation:

Step1: Analyze the limit as (t\to-\infty)

We know that for an exponential function (a^t) ((a>0,a\neq1)), when (a = \frac{1}{3}) and (t\to-\infty), let (u=-t), then (t\to-\infty) implies (u\to\infty). And ((\frac{1}{3})^t = 3^{-t}=3^{u}). [ \begin{align*} \lim_{t\to-\infty}f(t)&=\lim_{t\to-\infty}\left(-5\left(\frac{1}{3}\right)^t + 3\right)\ &=\lim_{u\to\infty}\left(-5\times3^{u}+3\right) \end{align*} ] As (u\to\infty), (3^{u}\to\infty), so (- 5\times3^{u}\to-\infty). Then (\lim_{t\to-\infty}f(t)=-\infty).

Step2: Analyze the limit as (t\to\infty)

For the exponential function (y = (\frac{1}{3})^t), since (0<\frac{1}{3}<1), we know that (\lim_{t\to\infty}(\frac{1}{3})^t = 0). [ \begin{align*} \lim_{t\to\infty}f(t)&=\lim_{t\to\infty}\left(-5\left(\frac{1}{3}\right)^t+3\right)\ &=-5\times\lim_{t\to\infty}\left(\frac{1}{3}\right)^t+\lim_{t\to\infty}3 \end{align*} ] Using the property of limits (\lim_{t\to\infty}c = c) (where (c) is a constant) and (\lim_{t\to\infty}a^t=0) for (0 < a<1). We get (-5\times0 + 3=3).

Answer:

As (t\to-\infty), (f(t)\to-\infty); As (t\to\infty), (f(t)\to3)