Below are some solutions to the sample final exam questions.

1. (a) L(x) = -3x + 7.     (b) f(x) = 16(1/4)^x.
2. y = -1/2 x + 3. Use implicit differentiation. Note that you don't have to waste time on excessive algebra to get a formula for dy/dx; just plug in x=0, and y=3 once you have differentiated implicitly, and then solve for dy/dx.
3. On the day before Christmas, it was snowing and the depth of snow on my lawn was increasing at a rate of 2 inches per day. However, it warmed up on Christmas morning and the snow on my lawn began melting at a rate of 3 inches per day. I must have had at least an inch of snow on my lawn on Dec. 24th since 3 - 2 = 1.
4. It's all about the Chain Rule.   (a) x e^{tan x} (2 + x sec²x)   (b) -2t²(t + 3)(t⁴ + 4t³)^-3/2   (c) -ln(2) 2^x sin(2^x)   (d) 1/[x(1 + (ln 5x)²)]
5. (a) f(x) is not differentiable at x = 1/2 (corner). (b) For f'(x), the graph will be zero at x = -1 and also between 1/2 and 2. It is negative (f is decreasing) for -2 < x < -1 and 2 < x < 4, and positive (f is increasing) for -1 < x < 1/2. For g'(x), note that g(x) looks very much like a shifted logarithmic function, (e.g., ln(x+ 0.6)), so the graph of g'(x) should look something like the graph of 1/x with x > 0.
6. -4/9. Use one of the two limit definitions of the derivative and simplify by adding the fractions in the numerator. Something should cancel (either h or x+3) before you can plug in to take the limit.
7. (a) 5/4. Factor and cancel, or use L'Hopital's Rule.   (b) -9/10. Use L'Hopital's Rule twice.   (c) -Infinity.   (d) Pi/4.
8. (a) There is a horizontal asymptote at y=0, both to the left and right. There are no vertical asymptotes; the denominator is always positive.   (b) f'(x) = (1 - x²)/(x² + 1)²   f''(x) = (2x(x² - 3))/(x² + 1)³. (c) There are two critical points at x = -1 and x = 1. The point (-1,-1/2) is a minimum while the point (1,1/2) is a maximum.   (d) There are three inflection points at (0, 0), (√3, √3/4), and (-√3, -√3/4).   (e) Click here for the graph. Note that f(x) is an odd function.
9. 2/3 by 5/3 by 20/3. If x represents the side length of the square removed from each corner, then the volume of the box is given by V(x) = x(8 - 2x)(3 - 2x). Expanding this expression out, computing V'(x) and solving V'(x) = 0 leads to a maximum at x = 2/3.
10. (a) p(x) = (-1/10)x + 550. The two data points on the price function are (1000, 450) and (1100, 440).   (b) $275. Compute R(x) = x p(x) and then solve R'(x) = 0.   (c) $350. Compute the profit function P(x) = R(x) - C(x), and then solve P'(x) = 0.
11. (a) True. (x(t))² + (y(t))² = 1 is satisfied for all t, so the curve traced out is the unit circle. It is traced out once from 0 to Pi/2, and then again from Pi/2 to Pi because the angle is 4t (not just t). You can plug in good t values like t = 0, Pi/8, Pi/4, etc. to see the direction the curve is being traced out.   (b) False. Consider f(x) = |x|, which is continuous at x = 0, but is not differentiable there because it has a corner.   (c) True.   (d) False. The acceleration is really 25e⁵ + 1.   (e) True.