College of the Holy Cross, Fall Semester, 2004 Math 131, Midterm 1 Solutions

1. [5 points each] The graph y = f(x) and four graphs obtained by transforming it are shown. Match the given formulas with the corresponding graph. Note that there is an extra graph.

−1  ^{0}  ^{1}  ^{2}  ^{3}_{−1}^{1}  y = f(x)
(a) y = f(2x): (iii)  (b) y = \frac{1}{2}f(x): (i)  (c) y = \frac{1}{2} − f(x): (ii)

Graph (iv) has equation y = f(\frac{1}{2}x).

−1  ^{0}  ^{1}  ^{2}  ^{3}_{−1}^{1}  (i)  ^{−1}  ^{0}  ^{1}  ^{2}  ^{3}_{−1}^{1}  (ii)

−1  ^{0}  ^{1}  ^{2}  ^{3}_{−1}^{1}  (iii)  ^{−2}  ^{0}  ^{2}  ^{4}  ^{6}_{−1}^{1}  (iv)

2. [20 points] The desert temperature H varies sinusoidally from a high of 80^{∘}F at 5 PM to a low of 40^{∘}F at 5 AM. Find a formula for H as a function of t, with t measured in hours from 5 PM. You may use the graph below for reference; it’s a good idea to start by labeling the vertical and horizontal scales.

0 20 40 60 80 100 5 PM  5 AM  5 PM Period = 24 hrs

Solution Because t = 0 at a “crest” (the temperature is maximum), we seek a function of the form H(t) = D + A cos(Bt) with A > 0 the amplitude of oscillation. The maximum minus minimum is 40 = 2A, so A = 20. The vertical shift D is the halfways point between the extremes, so D = \frac{1}{2}(80 + 40) = 60. Finally, the period is 24 hours, so B = 2π/24 = π/12. In summary

H(t) = 60 + 20 cos(\frac{πt}{12}).

3. [10 points each] In each part, fill in the table as indicated.

(a) Assuming f is a linear function.

Solution The line through the points (−1, 24) and (1, 12) has slope

m = (24 − 12)/(−1 − 1) = 12/ − 2 = −6.

The point-slope form of the equation is (y − 12) = −6(x − 1), which simplifies to y = 18 − 6x. To fill in the table, set x = 5 and x = 10:

\begin{matrix}x & −1 & 1 & 5 & 10 \\ f(x) & 24 & 12 & −12 & −42 \end{matrix}

(b) Assuming f is an exponential function.

Solution Here we are told to assume f(x) = P_{0}a^{x}. Using the values provided gives two equations,

24 = P_{0}a^{−1}, 12 = P_{0}a.

Dividing the first equation by the second cancels P_{0}: 2 = 24/12 = a^{−1}/a = a^{−2}, or a = 2^{−1/2}. Next, we plug this value into one of the equations (say the first) and solve for P_{0}: Since 24 = P_{0}a^{−1}, we have P_{0} = 24a = 24/\sqrt{2} = 12\sqrt{2}. Thus

f(x) = P_{0}a^{x} = (12\sqrt{2})2^{−x/2}.

As above, we fill in the table by setting x = 5 and x = 10.

\begin{matrix}x & −1 & 1 & 5 & 10 \\ f(x) & 24 & 12 & 3 & \frac{3\sqrt{2}}{8} ≃ 0.53 \end{matrix}

4. [10 points] Find the polynomial whose graph is shown; express your answer in both factored and expanded (multiplied out) form.

−2  −1  0  1  2  3  4 6 12 18  y = p(x)

Solution We have p(x) = k(x+1)^{2}(x−3). Since p(0) = 6, we find that 6 = k(1)^{2}(−3), or k = −2. Thus,

p(x) = −2(x + 1)^{2}(x − 3) (factored)
= −2(x^{2} + 2x + 1)(x − 3)
= −2(x^{3} − x^{2} − 5x − 3)
= −2x^{3} + 2x^{2} + 10x + 6. (expanded)

5. [10 points each] Let y = f(x) = x^{2} − 3 for x ≥ 0.

(a) Find the inverse function in the form y = f^{−1}(x), and find the domain of f^{−1}.

Solution Formally, exchange x and y, obtaining x = y^{2} − 3, then solve for y. This gives y = ±\sqrt{x + 3}. To see whether we need a negative sign or not, it’s easiest to consider the graph:

y = f(x) y = g(x)

We conclude that y = f^{−1}(x) = \sqrt{x + 3}.

(b) Suppose g(x) = x + e^{x}. With f(x) as above, find g(f(x)).

Solution g(f(x)) = g(x^{2} − 3) = (x^{2} − 3) + e^{(x^{2}−3)}.

6. [15 points] An automobile costs $25,000 and depreciates in value by 20% per year. How many years pass before the car is worth $5000? Give both an exact answer (using fractions, square roots, logarithms, etc. as needed) and a numerical answer rounded to two decimal places.

Solution Let V (t) be the value of the car in thousands of dollars after t years. (Note the explicit choice of units. This should be your first step when creating a mathematical model!) Because V (0) = 25 and V decreases to 0.8 times its current value after 1 year, we have

V (t) = 25(0.8)^{t}.

The problem asks for the value T so that V (T) = 25(0.8)^{T} = 5. Dividing by 25 and taking natural logarithms gives

(0.8)^{T} = 0.2,  T ln(0.8) = ln(0.2),  T = \frac{ln(0.2)}{ln(0.8)} ≃ 7.21 years