Question

1. a 12,000 - liter tank of water is filled to capacity. at time (t = 0), water begins to drain out of the tank at a rate modeled by (r(t)), measured in liters per hour, where (r) is given by the piece - wise defined function (r(t)=\begin{cases}\frac{600t}{t + 3},&0leq t<5\\1000e^{-0.2t},&t > 5end{cases}) (r(t)=\begin{cases}\frac{600t}{t + 3},&for 0leq tleq5\\1000e^{-0.2t},&for t>5end{cases}) (a) is (r) continuous at (t = 5)? show the work that leads to your answer. 2. given the function (f(x)=\frac{x^{3}+2x^{2}-3x}{3x^{2}+3x - 6}) (a) what are the zeros of (f(x))? (b) what are the vertical asymptotes of (f(x))? (c) what is (lim_{x

ightarrowinfty}f(x))?

Explanation:

Step1: Check left - hand limit of \(r(t)\) at \(t = 5\)

Calculate \(\lim_{t
ightarrow5^{-}}r(t)\). Since for \(0\leq t\leq5\), \(r(t)=\frac{600t}{t + 3}\), then \(\lim_{t
ightarrow5^{-}}\frac{600t}{t + 3}=\frac{600\times5}{5 + 3}=\frac{3000}{8}=375\)

Step2: Check right - hand limit of \(r(t)\) at \(t = 5\)

Calculate \(\lim_{t
ightarrow5^{+}}r(t)\). Since for \(t>5\), \(r(t)=1000e^{- 0.2t}\), then \(\lim_{t
ightarrow5^{+}}1000e^{-0.2t}=1000e^{-0.2\times5}=1000e^{-1}\approx1000\times0.3679 = 367.9
eq375\)
Since \(\lim_{t
ightarrow5^{-}}r(t)
eq\lim_{t
ightarrow5^{+}}r(t)\), \(r(t)\) is not continuous at \(t = 5\)

Step3: Find zeros of \(f(x)=\frac{x^{3}+2x^{2}-3x}{3x^{2}+3x - 6}\)

First, factor the numerator \(x^{3}+2x^{2}-3x=x(x^{2}+2x - 3)=x(x + 3)(x - 1)\) and the denominator \(3x^{2}+3x - 6=3(x^{2}+x - 2)=3(x + 2)(x - 1)\). The zeros of \(f(x)\) occur when the numerator is zero and the denominator is non - zero. Setting the numerator equal to zero gives \(x(x + 3)(x - 1)=0\). But \(x = 1\) makes the denominator zero, so the zero is \(x=0\) and \(x=-3\)

Step4: Find vertical asymptotes of \(f(x)\)

Vertical asymptotes occur when the denominator of \(f(x)\) is zero and the numerator is non - zero. Set \(3x^{2}+3x - 6 = 0\), which factors to \(3(x + 2)(x - 1)=0\). So the vertical asymptotes are \(x=-2\) and \(x = 1\)

Step5: Find \(\lim_{x

ightarrow\infty}f(x)\)
Divide both the numerator and denominator of \(f(x)=\frac{x^{3}+2x^{2}-3x}{3x^{2}+3x - 6}\) by \(x^{2}\). We get \(\lim_{x
ightarrow\infty}\frac{x^{3}+2x^{2}-3x}{3x^{2}+3x - 6}=\lim_{x
ightarrow\infty}\frac{x + 2-\frac{3}{x}}{3+\frac{3}{x}-\frac{6}{x^{2}}}=\infty\)

Answer:

(a) \(r(t)\) is not continuous at \(t = 5\)
(b) Zeros of \(f(x)\) are \(x = 0\) and \(x=-3\)
(c) Vertical asymptotes of \(f(x)\) are \(x=-2\) and \(x = 1\)
(d) \(\lim_{x
ightarrow\infty}f(x)=\infty\)