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Part 1. Solutions and Solution Sets
1. The width (shorter dimension) of the A0 paper is \(x\). When the paper is folded crosswise along the middle of its length (longer dimension) \(y\), the resulting halves of size A1 would have the same aspect ratio (width/length) as the original paper. Repeating the process of preserving the aspect ratio by folding results in the smaller sizes A2 (half of A1), A3 (half of A2), A4 (half of A3), and so on.
a. Using ratio and proportion, solve for the equation relating \(y\) and \(x\).
SOLUTION. Note that the aspect ratios of A0 and A1 are equal. Thus,
\[\begin{equation}\frac{x}{y} = \frac{y/2}{x}\end{equation}\]
By cross multiplication and taking the square root, we get:
\[\begin{equation}\boxed{y = \sqrt{2}x}\end{equation}\]
b. Solve for the aspect ratios (length \(\times\) width) in terms of \(x\) for paper sizes A0 to A4. [i.e. A0 is (\(\alpha_0 x \times x\)), A1 is (\(\alpha_1 x \times \beta_1 x\)), A2 is (\(\alpha_2 x \times \beta_2 x\)), and so on, where the \(\alpha\)’s and \(\beta\)’s are fractional values that you need to solve for.]
SOLUTION. We already obtained \(y\) as a function of \(x\), so we can now report everything in terms of \(x\).
A0 is \(\left(y \times x\right)\)) = \(\left(\sqrt{2}x \times x\right)\).
A1 is \(\left(x \times y/2\right)\) = \(\left(x \times \frac{\sqrt{2}}{2}x\right)\).
A2 is \(\left(y/2 \times x/2\right)\) = \(\left(\frac{\sqrt{2}}{2}x \times \frac{1}{2}x\right)\).
A3 is \(\left(x/2 \times y/4\right)\) = \(\left(\frac{1}{2}x \times \frac{\sqrt{2}}{4}x\right)\).
A4 is \(\left(y/4 \times x/4\right)\) = \(\left(\frac{\sqrt{2}}{4}x \times \frac{1}{4}x\right)\).
Part 2. Functions and Their Domains and Ranges
2. Consider the equations: \(y^2 = x - 6\) and \(y = (x - 6)^{1/2}\).
a. Which of these relations is/are function(s)? [You may draw the curves to justify your answer.]
SOLUTION. Only the second one is a function. For the first one, two values of \(y\) will be obtained for the same \(x\), e.g. \(x = 10\) will yield \(y = \pm 2\). The second one is practically the same as the first, except for the fact that we take only the positive square root of \(y\). This is observed from the plot below:
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| Plotted using desmos.com/calculator |
b. For the function(s), give the domain and range.
SOLUTION. From the figure, we see that the second equation has a domain of \(x \geq 6\) and range of \(y \in [0, \infty)\).
Part 3. Algebraic Operations and Function Inverses
Consider the functions: \(f(x) = 2x^3 - 3\) and \(g(x) = 1/(8x)\)
3. Do the ff. operations: (a) addition \(\left(f + g\right)(x)\); (b) subtraction \(\left(f - g\right)(x)\) and \(\left(g - f\right)(x)\); (c) multiplication \(\left(f \cdot g\right)(x)\); (d) division \(\left(f / g\right)(x)\) and \(\left(g/f\right)(x)\); and (e) composition \(\left(f \circ g\right)(x)\) and \(\left(g \circ f\right)(x)\).
SOLUTION.
a. \(\left(f + g\right)(x) = 2x^3 - 3 + \frac{1}{8x}\)
b. \(\left(f - g\right)(x) = 2x^3 - 3 - \frac{1}{8x} = -\left(g - f\right)(x)\)
c. \(\left(f \cdot g\right)(x) = \frac{1}{4}2x^2 - \frac{3}{8x}\)
d. \(\left(f / g\right)(x) = 16x^4 - 24x - \frac{1}{8x} = \frac{1}{\left(g - f\right)(x)}\)
e.1. \(\left(f \circ g\right)(x) = \frac{1}{256x^3} - 3\)
e.2. \(\left(g \circ f\right)(x) = \frac{1}{16x^3 - 24}\)
4. (a) Find the inverses of \(f(x)\) and \(g(x)\) and (b) show explicitly that both compositions result in \(x\).
SOLUTION.
a.1. \(f^{-1}(x) = \sqrt[3]{\frac{x + 3}{2}}\)
a.2. \(g^{-1}(x) = \frac{1}{8x}\)
b.1.1. \(f^{-1}\left(f(x)\right) = \sqrt[3]{\frac{\left(2x^3 - 3\right) + 3}{2}} = x\)
b.1.2. \(f\left(f^{-1}(x)\right) = 2\left[\sqrt[3]{\frac{x + 3}{2}}\right]^{3} - 3 = x\)
b.2. \(g^{-1}\left(g(x)\right) = \frac{1}{8\frac{1}{8x}} = g\left(g^{-1}(x)\right) = x\)
Part 4. Limits and Continuity
Consider the following functions: \(h(x) = \frac{2\sqrt{x} - 4}{8x - 32}\) and \(y(x)=\frac{x^2-25}{x^2+2x-15}\)
5. Solve for (a) \(\lim_{x \rightarrow 4} h(x)\) and (b) \(\lim_{x \rightarrow -5} y(x)\).
SOLUTION.
a. \(\lim_{x \rightarrow 4} \frac{2\sqrt{x} - 4}{8x - 32} = \lim_{x \rightarrow 4}\frac{2\sqrt{x} - 4}{2\left(2\sqrt{x} - 4\right)\left(2\sqrt{x} + 4\right)} = \lim_{x \rightarrow 4}\frac{1}{2\left(2\sqrt{x} + 4\right)} = \frac{1}{16}\)
b. \(\lim_{x \rightarrow -5} \frac{x^2-25}{x^2+2x-15} = \lim_{x \rightarrow -5} \frac{(x-5)(x+5)}{(x-3)(x+5)} = -\frac{5}{4} \)
6. Comment on the continuity (a) of \(h(x)\) at \(x = 4\) and (b) \(y(x)\) at \(x = -5\).
SOLUTION. Both are continuous at the given points because the limit is equal to the function evaluation at these points.
Part 5. The Derivative
7. Show, using the definition in terms of limits, that the derivative of \(R(x) = 2x^3 - 5x^2\) is \(R'(x) = 6x^2 – 10x\).
SOLUTION. Show.
\(R'(x) = \lim_{h \rightarrow 0} \frac{R(x+h) - R(x)}{h}\)
\(R'(x) = \lim_{h \rightarrow 0} \frac{2\left(x^3 +3x^2h + 3xh^2 + h^3\right) - 5\left(x^2 + 2xh + h^2\right) - \left(2x^3 - 5x^2\right)}{h}\)
\(R'(x) = \lim_{h \rightarrow 0} \frac{6x^2h + 6xh^2 + 2h^3 - 10xh - 5h^2}{h}\)
\(R'(x) = \lim_{h \rightarrow 0} \left(6x^2 + 6xh + 2h^2 - 10x - 5h\right)\)
\(R'(x) = 6x^2 - 10x\)
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