MATH 133 Calculus 1 with FUNdamentals
Sample Final Exam Solutions
Below are solutions to the sample final exam questions.
- (a) y = -3x + 7, (b) y = (1/3)x + 2
- (a) Domain: [-1, 1], Range: [0, π],
(b) cot θ = 3/4, csc θ = -5/4
(c) Period: 6π, Amplitude: 7
- (a) Domain: (-3, ∞), Range: All real numbers,
(b) x = -3
(c) x-intercept: e2 - 3 Graph
(d) f-1(x) = ex+2 - 3, Domain: All real numbers, Range: (-3, ∞)
- 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.
- (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.
- (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.
- 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.
- It's all about the Chain Rule!
(a) x esin-1 x (2 + x/sqrt{1 - x2})
(b) -2t2(t + 3)(t4 + 4t3)-3/2
(c) -ln(2) 2x sin(2x)
(d) 1/(x ln(5x) )
- (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.
- (a) L4 = 6.25 (overestimate),
(b) M4 = 5.375,
(c) Actual area is 16/3 = 5.333...
- (a) ½ e2,
(b) 1/3,
(c) 32 π
-
(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.