Calculus 1, MATH 135-08, 135-09, Fall 2017
Sample Final Exam Solutions
Below are some solutions to the sample final exam questions.
- (a) y = -3x + 7, (b) y = (1/3)x + 2
- (a) Domain: [-1, 1], Range: [π, 2π],
(c) Period: 6π, Amplitude: 7
- (a) Domain: (-3, ∞), Range: All real numbers,
(b) x = -3
(c) x-intercept: e2 - 3, y-intercept: ln(3) - 2
(d) f-1(x) = ex+2 - 3
- 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.
- (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.
- (a) 5/4. Factor and cancel, or use L'Hopital's Rule.
(b) 4/5. Use L'Hopital's Rule.
(c) -9/10. Use L'Hopital's Rule twice.
- 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 etan x (2 + x sec2x)
(b) -2t2(t + 3)(t4 + 4t3)-3/2
(c) -ln(2) 2x sin(2x)
(d) 1/[x(1 + (ln 5x)2)]
(a) f is an odd function since f(-x) = -f(x).
(b) There is a horizontal asymptote at y = 0, both to the left and right. There
are no vertical asymptotes; the denominator is always positive.
(c) f'(x) = (1 - x2)/(x2 + 1)2,
f''(x) = (2x(x2 - 3))/(x2 + 1)3.
(d) 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.
(e) There are three inflection points at (0, 0), (√3,
for the graph. Note that since f(x) is an odd function, its graph is symmetric with respect to the origin.
- 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.
(a) False. Consider f(x) = |x|, which is continuous at x = 0, but is not differentiable there because it has
(c) False. The acceleration is really 25e5 + 1.
(d) True. The curves intersect at the point (0, 1) with slopes 1 and -1, respectively.
(e) False. It is true that h'(π) = -3, but h''(π) = 5, not 2.