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Solution: APPM 1350 Exam 1 Spring 2011
- (30 pts) Use the limit definition of the derivative to find:
(a) f ′(x) given f (x) =
x^2 + 9
(b) f ′(0) given f (x) = 5x
4
(c)
d^2 f
dx^2
given
d
dx
f (x) =
10 x + 6
Solution:
(a) We have
f ′(x) = lim h→ 0
f (x + h) − f (x)
h
= lim h→ 0
(x + h)^2 + 9 −
x^2 + 9
h
(x + h)^2 + 9 +
x^2 + 9 √ (x + h)^2 + 9 +
x^2 + 9
= lim h→ 0
((x + h)^2 + 9) − (x^2 + 9)
h(
(x + h)^2 + 9 +
x^2 + 9)
= lim h→ 0
2 xh + h^2
h(
(x + h)^2 + 9 +
x^2 + 9)
2 x
2
x^2 + 9
x √ x^2 + 9
(b) We have
f ′(0) = lim x→ 0
f (x) − f (0)
x − 0
= lim x→ 0
5 x^4 + 3x + 20 − 20
x
= lim x→ 0
5 x^4 + 3x
x
= lim x→ 0
5 x^3 + 3 = 3
(c) We have
f ′′(x) = lim h→ 0
f ′(x + h) − f ′(x)
h
= lim h→ 0
1 10(x+h)+6 −^
1 10 x+
h
= lim h→ 0
10 x + 6 − (10(x + h) + 6)
(10(x + h) + 6)(10x + 6)
h
= lim h→ 0
− 10 h
(10(x + h) + 6)(10x + 6)
h
(10x + 6)^2
- Suppose the position function of a particle in motion is given by s(t) = 5t^4 + 3t + 20 ft, where t is in
seconds.
(a) (8 pts) Find the average rate of change of the position of the particle from t = 0 seconds to
t = 2 seconds.
(b) (6 pts) Find the instantaneous rate of change of the position of the particle when t = 0 seconds.
Solution:
(a) Here we have
Average Rate of Change =
s(2) − s(0)
2 − 0
= 43 ft/s
(b) From #1(b) we have s′(0) = 3 ft/s
- (18 pts) Let f (x) = x^2 + 1, m(x) = x and b(x) =
x^2 − 1,
(a) Find (f ◦ m ◦ b)(x) and state the domain.
(b) Find (f + m)(x) and state the domain.
(c) Find
b
m
(x) and state the domain.
Solution:
(a) (f ◦ m ◦ b)(x) = f (m(b(x))) = x^2 with domain (−∞, −1] ∪ [1, ∞)
(b) (f + m)(x) = x
2
- 1 + x with domain (−∞, ∞)
(c)
b
m
(x) =
x^2 − 1
x
with domain (−∞, −1] ∪ [1, ∞)
- (20 pts) Justify your answers for all the questions below. Consider the function f (x) =
x + 1
2 x
|x|
does f (x) have any:
(a) vertical asymptotes? If so, what are they?
(b) horizontal asymptotes? If so, what are they?
(c) removable discontinuities? If so, what are they?
(d) jump discontinuities? If so, what are they?
Solution:
(a) Note, lim x→− 1 −^
f (x) = −∞ and lim x→− 1 +^
f (x) = ∞, and so there is a vertical asymptote at x = − 1.
(b) Note, lim x→∞
f (x) = 2 and lim x→−∞
f (x) = −2, so the horizontal asymptotes are y = 2 and y = −2.
(c) No, there are NO removable discontinuities.
(d) Yes, there is a jump discontinuity at x = 0 since lim x→ 0 −^
f (x) = −1 and lim x→ 0 +^
f (x) = 3.
- (18 pts) Let a, b and c be constants and consider the function f (x) where
f (x) =
x + 6, x ≤ 0
cx^2 + bx + a, 0 < x < 1 7 x + c, x ≥ 1
(a) Find all values of a, b and c for which f (x) will be continuous at x = 0 and x = 1?
(b) For what values of a, b and c will f (x) be differentiable at x = 0 and x = 1?
(c) Find the equation of the tangent line to f (x) at x = −1.
Solution:
(a) For just continuity, we need a = 6 and b = 1 and c can have any value.
(b) Note that,
f ′(x) =
1 , x < 0 2 cx + 1, 0 < x < 1
7 , x > 1
and so for differentiability we need a = 6, b = 1 and c = 3.
(c) Here f ′(−1) = 1 and (− 1 , f (−1)) = (− 1 , 5) and so the equation is y = 5 + (x + 1) = x + 6.
