Math 1411: Exit Exam- Part 1 November 7, 2005

Problem 1. A culture bacteria originally numbers 500. After 2 hours there are 1500 bacteria. Assuming exponential growth, how many are there after 6 hours? What is the doubling time of this bacteria?

Answer. If $f(t)$ measures the bacteria at time $t$, then we have

$$f(t) = A e^{k\ t}$$

where $t$ is measured in hours. Since $A$ is the initial amount, we must have $A = 500$, or $f(t) = 500 e^{k\ t}$. Since $f(2) = 1500$ we get

$$1500 = 500 e^{k \ 2}$$

or $e^{2k} = 3$ which implies $2 k = \ln(3)$ or

$$k = \frac{\ln(3)}{2} = 0.549$$

After 6 hours, we have 13475 bacteria. For the doubling time $T$, we have $f(T) = 2 \times 500 = 1000$. So

$$500 e^{k \ T} = 1000$$

or $e^{k \ T} = 2$ which implies $k \ T = \ln(2)$ or

$$T = \frac{\ln(2)}{k} = 1.261$$

or $T$ is 1 hour 15 minutes and 42 seconds...

Problem 2. Find the amplitude and period of $f(x) = 4-2\cos(5x)$. Then explain how to graph it using the graph of $h(x) = \cos(x)$.

Answer. The maximum value is $\max = 4+2 = 6$ and the minimum value is $\min = 4-2 = 2$. So the middle value is $$\frac{6+2}{2} = 4$$. So the amplitude is $6-4 = 2$. The period is $$\frac{2 \pi}{5}$$. For the graph we have

$$\begin{array}{ccllll} \cos(x) & \rightarrow & \cos(5x) &\mbox... ...\hspace*{1cm}}& \mbox{Vertical Shift up 4 units}\\ \end{array}$$

Problem 3. Investigate $$\lim_{h \rightarrow 0} (1+h)^{1/h}$$ numerically.

Answer. We have

$$\begin{array}{llll} h & \mbox{\hspace*{2cm}} & (1+h)^{1/h} \\... ...\\ 0.000001 &\mbox{\hspace*{2cm}} & 2.718281378\\ \end{array}$$

So it looks like that we have

$$\lim_{h \rightarrow 0} (1+h)^{1/h} = 2.718281$$

In fact we have

$$(1+h)^{1/h} = e^{\displaystyle \frac{\ln(1+h)}{h}}$$

and since

$$\lim_{h \rightarrow 0} \frac{\ln(1+h)}{h} = 1$$

we get

$$\lim_{h \rightarrow 0} (1+h)^{1/h} = e^{1} = e = 2.718281828..$$

Problem 4. Find a value of the constant $k$ such that the limit exists

(a)

$$\lim_{x \rightarrow 4} \frac{x^2-k^2}{x-4}$$

(b)

$$\lim_{x \rightarrow 1} \frac{x^2-kx + 4}{x-1}$$

Answer. For (a), we have

$$\lim_{x \rightarrow 4} \frac{x^2 - k^2}{x-4} = \frac{16-k^2}{4-4}$$

If $k^2 \neq 16$, then the limit does not exist. So we must have $k^2 = 16$. In this case we have

$$\lim_{x \rightarrow 4} \frac{x^2 - 16}{x-4} = \lim_{x \rightarrow 4} \frac{(x-4)(x+4)}{x-4} =\lim_{x \rightarrow 4} x+4 = 8$$

For (b) we have

$$\lim_{x \rightarrow 1} \frac{x^2 - kx + 4}{x-1} = \frac{1 - k + 4}{1-1}$$

If $k \neq 5$, the limit does not exist. So we must have $k = 5$. In this case we have

$$\lim_{x \rightarrow 1} \frac{x^2 - 5x + 4}{x-1} = \lim_{x \ri... ... \frac{(x-1)(x-4)}{x-1} = \lim_{x \rightarrow 1} x-4 = 1-4 = -3$$

Problem 5. Find the constant $A$ which makes $f(x)$ continuous at $x=2$ where

$$f(x) = \left\{\begin{array}{lll} x^2-3 & x < 2 \\ x+A & x \geq 2 \end{array} \right.$$

Answer. Because $f(x)$ is a piecewise function, then we will compute the left-limit and right-limit of $f(x)$ at $x=2$. We have

$$\lim_{x \rightarrow 2+} f(x) = \lim_{x \rightarrow 2+} x+A = 2+A$$

and

$$\lim_{x \rightarrow 2-} f(x) = \lim_{x \rightarrow 2-} x^2-3 = 4-3 = 1$$

So if $f(x)$ is continuous at $x=2$ its limit at $x=2$ exists. In this case the left-limit and the right-limits are equal which implies

$$A+2 = 1$$

or $A = -1$. In this case we have

$$\lim_{x \rightarrow 2} f(x) = 1 = f(2)$$

which means that $f(x)$ is continuous at $x=2$.

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