MATH 121 (9am, S.Smith) Exam 2 Solutions Fri 15 Oct 2004

1. (10 pts.) Find the remainder when f(x) = x^{10} + 2x^{8} is divided by x + 1. (Hint: you do not need to actually divide.) By Remainder Theorem, get f(−1) = (−1)^{10} + 2(−1)^{8} = 1 + 2 = 3.

2. (15 pts.) Find a polynomial f(x) with real coefficients, of degree 4, whose only roots are 4, 3 + i, and 3 − i. Complex roots occur in conjugate pairs, so 4 must have multiplicity 2. So (x − 4)^{2}(x − (3 + i))(x − (3 − i)) works (can multiply by any constant). This gives (x^{2} − 8x + 16)(x^{2} − 6x + 10) = x^{4} − 14x^{3} + 74x^{2} − 176x + 160.

3. (15 pts.) Given the rational function \frac{2 − x}{x − 3}, find the x- and y-intercepts; find the vertical and horizontal asymptotes; and sketch a graph, labeling those features. For x = 0, the y-intercept is the function value \frac{2}{−3} = −\frac{2}{3}. For y = 0, the x-intercept is the root 2 of the numerator 2 − x. The vertical asymptote is the root x = 3 of the denominator x − 3. For the horizontal asymptote, divide numerator and denominator by x,

to get \frac{\frac{2}{x} − 1}{1 − \frac{3}{x}}; for large values of |x|, this approaches \frac{−1}{1} = −1.

4. (15 pts.) Factor f(x) = x^{3} + x^{2} + x + 1 completely over the complex numbers, given that one root is i. Then −i is also a root, so f(x) is divisible by (x + i)(x − i) = x^{2} + 1. The quotient is x + 1, so f(x) = (x + i)(x − i)(x + 1).

5. (15 pts.) Given f(x) = e^{2x}, compute the difference quotient \frac{f(x + h) − f(x)}{h}, and simplify (assuming h ≠ 0).

\frac{e^{2(x+h)} − e^{2x}}{h} = \frac{e^{2x}e^{2h} − e^{2x}}{h} = e^{2x} · \frac{e^{2h} − 1}{h}.

6. (15 pts.) How long will it take to double an investment of $ 500 at 7 % annual interest, compounded continuously? 1000 = 500e^{.07t}, so 2 = e^{.07t}, so ln(2) = .07t, and then t = \frac{ln(2)}{.07} ≅ 9.9021 years.

7. (15 pts.) Solve ln(3x − 5) = ln(11) + ln(2). Exponentiate both sides: 3x − 5 = 11 · 2 = 22, so 3x = 27 and hence x = 9.