Below are solutions to the sample final exam questions.

1. (a) y = -3x + 7,     (b) y = (1/3)x + 2
2. (a) Domain: [-1, 1], Range: [0, π],     (b) cot θ = 3/4, csc θ = -5/4     (c) Period: 6π, Amplitude: 7
3. (a) Domain: (-3, ∞), Range: All real numbers,     (b) x = -3     (c) x-intercept: e² - 3     Graph     (d) f^-1(x) = e^x+2 - 3,   Domain: All real numbers, Range: (-3, ∞)
4. y = -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=3, and y=0 once you have differentiated implicitly, and then solve for dy/dx.
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 or 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. (a) -1. Plug in x = π and simplify.   (b) 5/4. Use L'Hopital's Rule or factor and cancel.   (c) 4/5. Use L'Hopital's Rule.   (d) 5. Use L'Hopital's Rule twice.   (e) π/4.
7. 1/2. Use one of the two limit definitions of the derivative and simplify by multiplying top and bottom by the conjugate. Something should cancel (either h or x-3) before you can plug in and evaluate the limit.
8. It's all about the Chain Rule!   (a) x e^{sin-1 x} (2 + x/sqrt{1 - x²})   (b) -2t²(t + 3)(t⁴ + 4t³)^-3/2   (c) -ln(2) 2^x sin(2^x)   (d) 1/(x ln(5x) )
9. (a) There is a horizontal asymptote at y = 0, but there are no vertical asymptotes because the denominator is always positive.
   (b) 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.
10. (a) L₄ = 6.25 (overestimate),     (b) M₄ = 5.375,     (c) Actual area is 16/3 = 5.333...
11. (a) ½ e²,     (b) 1/3,     (c) 32 π
12. (a) False. Consider f(x) = |x|, which is continuous at x = 0, but is not differentiable there because it has a corner.
    (b) True.
    (c) False, since h'(π) = -3.
    (d) False. The value of the integral is 8 using linearity and the fact that the integral of an odd function over a symmetric interval is zero.