Calculus 1, MATH 13508, 13509, 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],
(b) 3/4,
(c) Period: 6π, Amplitude: 7
 (a) Domain: (3, ∞), Range: All real numbers,
(b) x = 3
(c) xintercept: e^{2}  3, yintercept: ln(3)  2
(d) f^{1}(x) = e^{x+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.
(d) Infinity.
(e) Pi/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 x3) before you can plug in and evaluate the limit.
 It's all about the Chain Rule.
(a) x e^{tan x} (2 + x sec^{2}x)
(b) 2t^{2}(t + 3)(t^{4} + 4t^{3})^{3/2}
(c) ln(2) 2^{x} sin(2^{x})
(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  x^{2})/(x^{2} + 1)^{2},
f''(x) = (2x(x^{2}  3))/(x^{2} + 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,
√3/4), and
(√3,
√3/4).
(f) Click
here
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
a corner.
(b) True.
(c) False. The acceleration is really 25e^{5} + 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.