Calculus 1 with FUNdamentals, MATH 13302
Sample Final Exam Solutions
Below are some solutions to the sample final exam questions.
 (a) L(x) = 3x + 7. (b) f(x) = 16(1/4)^{x}.
 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.
 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.
 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(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.
 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.
 (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.

(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  x^{2})/(x^{2} + 1)^{2}
f''(x) = (2x(x^{2}  3))/(x^{2} + 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.
 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) 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.

(a) False. sin^{2}(3t) + cos^{2}(3t) = 1 for any real number t.
(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 25e^{5} + 1.
(e) True.