College of the Holy Cross, Fall 2013 Math 135, Section 1, Midterm 1 Solutions Friday, September 20

I. Match the plots below with the following formulas. Note that there is an extra plot.

(5) A) y = 3 − (x − 1)^{2} is Plot: \underline{3} (the x − 1 shifts the parabola to the right, not the left)

(5) B) y = sin(x/2) is Plot: \underline{2} (the x/2 makes the period equal to 4π ≐ 12.6, so the usual sine graph is stretched horizontally)

(5) C) y = 1 − e^{−x} is Plot: \underline{1} (think: y = e^{x} reflected across the x- and y-axes, then shifted up)

(5) D) y = sin(2x) is Plot: \underline{5} (the 2x makes the period equal to π ≐ 3.14, so the usual sine graph is compressed horizontally).

Plot 1: Plot 2:

Plot 3: Plot 4:

Plot 5:

II. The manager of a furniture factory has collected the following data for the cost of manufacturing chairs.

# Chairs (per day) x Cost (in dollars) y
100  2400
150  3100
250  4500
300  5200

(10) A) Given that y is a linear function of x, determine a formula for it.

The slope is m = \frac{3100−2400}{150−100} = 14 so by the point slope form, we get y − 2400 = 14(x − 100), or y = 14x + 1000.

Cost function: y − 2400 = 14(x − 100) or y = 14x + 1000

(5) B) What does the slope represent in real-world terms?

The slope represents the cost of manufacturing one additional chair per day, or C(x + 1) − C(x).

(5) C) What does the y-intercept represent in terms of cost?

The y-intercept of 1000 represents the cost per day if no chairs are actually manufactured (x = 0). These are often called fixed costs – things like the maintenance costs of the factory, taxes, labor costs, etc.

(5) D) Using your model, determine how much it will cost to produce 350 chairs per day.

Cost: (14)(350) + 1000 = $ 5900

III. Given f(x) = 4 − x^{2} and g(x) = \sqrt{3x − 2}, answer the following questions.

(10) A) Find the domain of f(x) and the domain of g(x).

The domains here are the sets of all real x that can be substituted into the formulas to yield a well-defined result. For f there are no restrictions. For g, we must have 3x − 2 ≥ 0, so x ≥ \frac{2}{3}.

Domain of f: all real x, or (−∞, +∞)

Domain of g: all real x ≥ \frac{2}{3}, or [\frac{2}{3}, +∞)

(5) B) What is the domain of the function g(x)/f(x)?

Now we must be able to substitute an x that makes sense for g, and that also avoids making f(x) = 0.

Domain of g(x)/f(x): [\frac{2}{3}, 2) ∪ (2, +∞), or something equivalent

(5) C) Find the function g ∘ f.  (g ∘ f)(x) = g(f(x)) = \sqrt{3(4 − x^{2}) − 2} = \sqrt{10 − 3x^{2}}

IV. Answer the following questions.

(5) A) Find all values of x in (−\frac{π}{2} , \frac{π}{2} ) for which |tan x| > 1.

This is true if tan(x) > 1 or tan(x) < −1. The first occurs for x between \frac{π}{4} and \frac{π}{2}; the second occurs for x between −\frac{π}{2} and −\frac{π}{4}.

Values of x: (−\frac{π}{2}, −\frac{π}{4}) ∪ (\frac{π}{4}, \frac{π}{2})

Note: Equivalent answers like: all x with −\frac{π}{4} < x < −\frac{π}{4} or \frac{π}{4} < x < \frac{π}{2} are also OK.

(5) B) If sin θ = \frac{2}{3} and \frac{π}{2} < θ < π, give the exact value of cos θ.

We can use the basic trig identity sin^{2}(θ) + cos^{2}(θ) = 1 for this: (\frac{2}{3})^{2} + cos^{2}(θ) = 1 so cos^{2}(θ) = \frac{5}{9} and cos(θ) = ±\frac{\sqrt{5}}{3}. Since θ is between \frac{π}{2} and π, the cosine must be negative, so the correct answer is:

cos θ: − \frac{\sqrt{5}}{3}

(5) C) Express as a single logarithm: \frac{1}{2} ln 3 − 3 ln 2 + ln 6.

Use the properties of logarithms: ln(A) + ln(B) = ln(AB), ln(A) − ln(B) = ln(A/B), and p ln(A) = ln(A^{p}). Then

\frac{1}{2} ln 3 − 3 ln 2 + ln 6 = ln (\frac{3^{1/2} · 6}{2^{3}}) .

Single logarithm: ln (\frac{3^{1/2}·6}{2^{3}}) = ln (\frac{3\sqrt{3}}{4})

V. Consider the function f(x) = \frac{1}{2}e^{x+1} + 1.

(15) A) Given that f is one-to-one, find a formula for the inverse function of f.

Set up y = \frac{1}{2}e^{x+1} + 1 and solve for x:

2(y − 1) = e^{x+1}, so after taking natural log of both sides
ln(2(y − 1)) = x + 1
x = ln(2(y − 1)) − 1.

We can swap the variables to write the inverse function as a function of x:

f^{−1}(x) = ln(2(x − 1)) − 1

(10) B) In the space below, plot the graphs of the functions f and f^{−1} on the same set of axes. Label one point on each graph with its coordinates.

Here are the graphs:

Note that f(x) > 1 for all x. This means that y = f(x) should be approaching the horizontal line y = 1 as x → −∞. Because of this, the graph y = f^{−1}(x) has a vertical asymptote at x = 1. It is obtained by reflecting y = f(x) across the line y = x. The top (red) curve is y = f(x); it contains the point (0, \frac{e}{2} + 1) ≐ (0, 2.36). The bottom (blue) curve is y = f^{−1}(x), obtained by reflecting y = f(x) across the line y = x; it contains the point (\frac{e}{2} + 1, 0) ≐ (2.36, 0).