Solution: APPM 1350
Exam #1 Review Questions
Summer 2014
1. Consider f (x) = x3 + 2x2 and g(x) = 3x2 − 1. Find f − g and 2f /g and state their domains.
Solution:
(f − g)(x) = x3 − x2 + 1 with all real numbers as its domain, and (2f /g)(x) =
all real x such that x 6= ± √13 .
2. If f (x) =
√
2x3 + 4x2
with domain
3x2 − 1
2x + 3 and g(x) = x2 + 5, find f ◦ g and f ◦ f and state their domains.
Solution:
p
√
Note that (f ◦ g)(x) = f (g(x))√= 2(x2 + 5) + 3 = 2x2 + 13 with all real numbers as its domain,
and (f ◦ f )(x) = f (f (x)) = (2 2x + 3 + 3)1/2 with domain x ≥ −3/2.
3. Evaluate the limit:
cos4 (x)
x2 − 4x
(a) lim
(b)
lim
x→∞ 5 + x2
x→4 x2 − 3x − 4
2x + 12
x5/3
(e) lim
(f) lim 5/3
x→∞ x
x→−6 |x + 6|
+1
2−x
x2 − 2x + 3
(i) lim
(j) lim
x→∞
x→1 x − 1
5 − 2x3
(x + h)−1 − x−1
h→0
h
(c) lim
(g) lim x2 − x4
x→∞
Solution:
(a) Note that 0 ≤ cos4 (x) ≤ 1 implies 0 ≤
sin2 (3t)
t→0
t2
3x + 2
(h) lim √
x→−∞
9x2 + 1
(d) lim
1
cos4 (x)
≤
, and so,
5 + x2
5 + x2
cos4 (x)
1
≤ lim
=0
x→∞ 5 + x2
x→∞ 5 + x2
0 ≤ lim
cos4 (x)
= 0.
x→∞ 5 + x2
thus by the Squeeze Theorem, lim
(b)
x2 − 4x “0/0”
x(x − 4)
x
= lim
= lim
= 4/5
x→4 x2 − 3x − 4
x→4 (x + 1)(x − 4)
x→4 (x + 1)
lim
(c)
(x + h)−1 − x−1 “0/0”
= lim
h→0
h→0
h
lim
1
x+h
−
h
1
x
= lim
h→0
x − (x + h) 1
−1
1
· = lim
=− 2
(x + h)x
h h→0 (x + h)x
x
(d)
sin2 (3t) “0/0”
9 sin(3t) sin(3t)
·
=9·1·1=9
= lim ·
2
t→0
t→0 1
t
3t
3t
2(x + 6)
2(x + 6)
x + 6, if x ≥ −6
(e) Recall that |x+6| =
thus, lim
= 2 and lim
= −2
+
−
−(x + 6), if x < −6
x→−6 (x + 6)
x→−6 −(x + 6)
2x + 12
and so lim
= does not exist.
x→−6 |x + 6|
lim
(f)
x5/3
x5/3
=
lim
x→∞ x5/3 + 1
x→∞ 5/3
x
1+
lim
1
1
1 =1
1 + x5/3
= lim
x→∞
x5/3
(g) lim x2 − x4 = lim x2 (1 − x2 ) = −∞
x→∞
x→∞
(h)
2
3x(1 + 3x
)
q
x→−∞
3|x| 1 + 9x12
x2 =|x|
3x + 2
=
lim √
x→−∞
9x2 + 1 √|{z}
(i) Note lim
x→1−
lim
2
3x(1 + 3x
)
q
= −1
x→−∞
1
−3x
1
+
x<0 so |x|=−x
9x2
=
|{z}
lim
2−x
2−x
2−x
= −∞ and lim
= ∞ and so lim
= does not exist.
x→1 x − 1
x−1
x→1+ x − 1
(j) Since the degree of the denominator is higher than the degree of the numerator, by dominance of
x2 − 2x + 3
powers we have, lim
= 0.
x→∞
5 − 2x3
4. If f (x) = x2 + 10 sin(x) show that there is a number c such that f (c) = 1000.
Solution: Note that −1 ≤ sin(x) ≤ 1 and so,
f (30) = 900 + 10 sin(30) ≤ 910 < 1000 and f (40) = 1600 + 10 sin(40) ≥ 1590 > 1000,
that is, f (30) < 1000 < f (40) and since f (x) is continuous for all real x, by the Intermediate Value
Theorem there exists a number c such that 30 < c < 40 and f (c) = 1000.
Alternatively: Set g(x) = x2 + 10 sin x − 1000 and show that there is a number c at which g(c) = 0.
Notice that
g(−10) = 100 + 10 sin (−10) − 1000 = −900 + 10 sin (−1) < 0,
and
g(100) = 10000 + 10 sin (100) − 1000 = 9000 + 10 sin (100) > 0.
Since f (x) is continuous everywhere, by the IVT, there is a number c such that −10 < c < 100 and
g(c) = 0 (that is, f (c) = 1000).
5. Find all values of a and b that make f continuous everywhere

x + 2, if x < 0

2ax2 + b, if 0 ≤ x ≤ 1
f (x) =

2 − x, if x > 1
Sketch the graph.
Solution:
Note, at x = 0 we have lim f (x) = lim x + 2 = 2 and limx→0+ f (x) = limx→0+ 2ax2 + b = b, so
x→0−
x→0−
b = 2 and at x = 1 we have lim f (x) = lim 2ax2 +2 = 2a+2 and limx→1+ f (x) = limx→1+ 2−x = 1
x→1−
x→1−
so we need 2a + 2 = 1 and so a =
− 1/2
thus we have

x + 2, if x < 0

−x2 + 2, if 0 ≤ x ≤ 1
f (x) =

2 − x, if x > 1
and the graph looks like
3
y=-x2+2
2
y=x+2
1
y=2-x
-4
-3.2
-2.4
-1.6
-0.8
0
0.8
1.6
2.4
3.2
4
-1
-2
-3
6. Find and classify the discontinuities of the given functions as either jump, infinite or removable:
(a) f (x) =
x2 − 2x − 8
x+2
(b) f (x) =
x−7
|x − 7|
(c) f (x) =
x − 90
x − 16x + 60
Solution:
x2 − 2x − 8
(x + 2)(x − 4)
= lim
= −6, but f (−2) is undefined, so there is a removx→−2
x→−2
x+2
x+2
able discontinuity at x = −2.
(a) Note lim
(b) Here, lim f (x) = −1 and lim f (x) = 1, so there is a jump discontinuity at x = 7.
x→7−
(c) Note, f (x) =
x→7+
x − 90
x − 90
x − 90
=
=
, so there is an infinite discontinuity at
x − 16x + 60
−15x + 60
−15(x − 4)
x = 4.
√
7. Find all horizontal and vertical asymptotes of f (x) =
4x2 + 1
.
2x − 5
Solution: Note that there is a vertical asymptote at x = 5/2 and lim f (x) = ∞ and
x→5/2+
lim f (x) = −∞.
x→5/2−
The horizontal asymptotes are y = 1 and y = −1, since
q
q
q
√
1
1
1
+
1
+
1 + 4x12
2|x|
2x
4x2 + 1
4x2
4x2
lim
=
lim
=
lim
= lim
=1
|{z}
x→∞ 2x − 5 √|{z} x→∞ 2x(1 − 5 )
x→∞ 2x(1 − 5 )
x→∞ 1 − 5
2x
2x
2x
x>0 so |x|=x
x2 =|x|
and,
√
lim
x→−∞
4x2
2|x|
q
1+
+1
= lim
x→−∞ 2x(1 −
2x − 5
1
4x2
=
lim
5
|{z}
x→−∞
2x ) x<0 so |x|=−x
−2x
q
1+
2x 1 −
q
− 1+
1
4x2
5
2x
= lim
x→∞
1−
1
4x2
5
2x
= −1
8. The displacement (in meters) of a particle moving in a straight line is given by s(t) = t2 − 8t + 18,
where t is measured in seconds.
(a) Find the average velocity of the particle over the time interval (i) [3, 4] and (ii) [4, 5].
(b) Find the instantaneous velocity when t = 4.
(c) When is the particle moving forward? Backward?
(d) Find the acceleration of the particle at time t = 4.
Solution:
(a) We have (i)
s(4) − s(3)
s(5) − s(4)
= 2 − 3 = −1 m/s and (ii)
= 3 − 2 = 1 m/s.
4−3
5−4
s(t) − s(4)
t2 − 8t + 18 − 2
t2 − 8t + 16
(t − 4)2
= lim
= lim
= lim
= lim(t − 4) = 0.
t→4
t→4
t→4
t→4 t − 4
t→4
t−4
t−4
t−4
(b) s0 (4) = lim
(c) The velocity function of the particle is v(t) = s0 (t) = 2t − 8 = 2(t − 4). Velocity is negative when
0 ≤ t < 4; velocity is positive when t > 4. Consequently, the particle moves forward when t is in
(4, ∞), and the particle moves backward when t is in [0, 4).
(d) The acceleration function of the particle is a(t) = v 0 (t) = s00 (t) = 2. Thus at t = 4 (and at any
t-value), the acceleration is 2 m/s2 .
9. Find an equation of the tangent line to the curve y = 2x(x + 1)2 at the point x = 1.
Solution:
Let f (x) = y = 2x(x + 1)2 . First, notice that when x = 1, we have y = f (1) = 8. Consequently, we
want the equation of the line that passes through the point (1, 8) with slope,
f (1 + h) − f (1)
2(1 + h)(1 + h + 1)2 − 8
(1 + h)(2 + h)2 − 4
= lim
= 2 lim
h→0
h→0
h→0
h
h
h
2
2
2
3
(1 + h)(4 + 4h + h ) − 4
4 + 4h + h + 4h + 4h + h − 4
= 2 lim
= 2 lim
h→0
h→0
h
h
h3 + 5h2 + 8h
= 2 lim
= 2 lim h2 + 5h + 8 = 2 · (0 + 0 + 8) = 16.
h→0
h→0
h
m = f 0 (1) = lim
The equation of the tangent line is y − f (1) = f 0 (1)(x − 1), so
y − 8 = 16(x − 1)
or y = 16x − 8.
10. Use the limit definition of the derivative to find f 0 (x) given:
(a) f (x) =
√
2x
(b) f (x) = x2 − 2x + 1
(c) f (x) =
1
x
Solution: (a)
f (x + h) − f (x)
f (x) = lim
h→0
h
0
p
p
p
√
√
√
2(x + h) − 2x
2(x + h) − 2x
2(x + h) + 2x
√
= lim
= lim
·p
h→0
h→0
h
h
2(x + h) + 2x
2
2(x + h) − 2x
1
= lim p
√ =√
= lim p
√
h→0
h→0
2x
2(x + h) + 2x
h
2(x + h) + 2x
(b)
f (x + h) − f (x)
h→0
h
f 0 (x) = lim
(x + h)2 − 2(x + h) + 1 − (x2 − 2x + 1)
h→0
h
2xh + h2 − 2h
= lim
= lim (2x + h − 2) = 2x − 2
h→0
h→0
h
=
lim
(c)
f (x + h) − f (x)
f (x) = lim
= lim
h→0
h→0
h
0
1
x+h
−
h
1
x
x − (x + h) 1
−1
1
· = lim
=− 2
h→0 (x + h)x
h h→0 (x + h)x
x
= lim
11. Use the limit definition of the derivative to find f 00 (x) given f (x) = x2 .
Solution: First we need to find f 0 (x),
f (x + h) − f (x)
(x + h)2 − x2
2xh + h2
= lim
= lim
= lim 2x + h = 2x
h→0
h→0
h→0
h→0
h
h
h
f 0 (x) = lim
then,
f 0 (x + h) − f 0 (x)
2(x + h) − 2x
2h
= lim
= lim
= 2.
h→0
h→0
h→0 h
h
h
f 00 (x) = lim