Mathematics 132 First Test
SOLUTIONS
13 June 2012
1. (20 points) Evaluate each limit.
(a) lim (2x2 − 4x + 5)
x→3
Solution:
lim (2x2 − 4x + 5) = 2(3)2 − 4(3) + 5 = 18 − 12 + 5 = 11.
x→3
x2 + x − 2
x→−2
x+2
Solution:
(b) lim
(x + 2)(x − 1)
x2 + x − 2
= lim
= lim (x − 1) = −2 − 1 = −3.
x→−2
x→−2
x→−2
x+2
x+2
lim
x−1
x2 − 1
Solution:
(c) lim
x→1
lim
x→1
x−1
x−1
1
1
1
= lim
= lim
=
= .
x2 − 1 x→1 (x + 1)(x − 1) x→1 x + 1
1+1
2
x2 − 3x + 2
x→∞ 4x2 + 6x − 10
Solution:
(d) lim
3
1− +
x2 − 3x + 2
x
lim
= lim
x→∞ 4x2 + 6x − 10
x→∞
6
4+ −
x
2
x2 = 1 − 0 + 2 = 1 .
4
10 4 + 0 − 0
2
x
x2 − 9
x→∞ x3 − 4x2 + 1
Solution:
(e) lim
x2 − 9
lim 3
= lim
x→∞ x − 4x2 + 1
x→∞
1
9
− 3
0
0−0
x x
= = 0.
=
1
−
0
+
0
1
4
1
1− + 3
x x
2. (30 points) Find the derivative f 0 for each function f . You are not required to simplify your answer.
(a) f (x) = x2012
Solution: f 0 (x) = 2012x2011
(b) f (x) = 211x
Solution: f 0 (x) = 211
(c) f (x) = 5x2 − 8x + 80
Solution: f 0 (x) = 10x − 8
Ogül Arslan/Eric Fu
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Overleaf. . .
(d) f (x) =
√
5
1
x = x5
Solution: f 0 (x) =
1 −4
x 5
5
(e) f (x) = e2
Solution: f 0 (x) = 0
1
= x−4
x4
Solution: f 0 (x) = −4x−5
√
1
(g) f (x) = x2 + 4x − 3 = (x2 + 4x − 3) 2
(f) f (x) =
Solution: f 0 (x) =
1 2
1
(x + 4x − 3)− 2 (2x + 4)
2
(h) f (x) = (x3 − 9x2 + x − 2)8
Solution: f 0 (x) = 8(x3 − 9x2 + x − 2)7 (3x2 − 18x + 1)
(i) f (x) =
x2 + 10x − 8
6x − 13
Solution: f 0 (x) =
(2x + 10)(6x − 13) − (x2 + 10x − 8)(6)
(6x − 13)2
(j) f (x) = (x3 − 5x) · (x4 + 5x2 )
Solution: f 0 (x) = (3x2 − 5)(x4 + 5x2 ) + (x3 − 5x)(4x3 + 10x)
3. (12 points) Find the equation of the tangent line to the graph of f (x) = 2x2 − 3x + 4 at (2, 6).
Solution: The first derivative is f 0 (x) = 4x − 3 and the slope of the tangent line at (2, 6) is
m = f 0 (2) = 4(2) − 3 = 5.
Using the point-slope formula, the equation of the tangent line is
y − 6 = 5(x − 2)
y − 6 = 5x − 10
y = 5x − 4.
4. (12 points) Use implicit differentiation to find
dy
for xy + 2x3 + y 2 = 10.
dx
Solution:
y+x
Ogül Arslan/Eric Fu
dy
dy
+ 6x2 + 2y
dx
dx
dy
dy
x
+ 2y
dx
dx
dy
(x + 2y)
dx
dy
dx
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=0
= −6x2 − y
= −(6x2 + y)
=−
6x2 + y
.
x + 2y
Overleaf. . .
5. The graph of a function f is given.
4
3
y = f (x)
2
1
−3
−2
−1
0
1
2
−1
(a) (6 points) Find
i.
lim f (x)
ii.
x→−1−
lim f (x)
iii. f (−1)
x→−1+
Solution:
lim f (x) = 2,
lim f (x) = 1,
x→−1−
x→−1+
f (−1) = 1.
(b) (6 points) Is the function f continuous? Justify your answer using the definition of continuity.
Solution: No, the function f is not continuous at x = −1. This is because the overall limit
lim f (x) does not exist, for lim f (x) 6= lim f (x).
x→−1−
x→−1
x→−1+
6. The demand function for the electronic speaker systems manufactured by the Weierstrass Company is
p = −0.04x + 800,
0 ≤ x ≤ 20, 000
where p denotes the unit price of the speakers (in USD) and x denotes the quantity demanded.
(a) (4 points) Find the revenue function R.
Solution:
R(x) = p · x = (−0.04x + 800)(x) = −0.04x2 + 800x USD.
(b) (4 points) Find the marginal revenue function R0 .
Solution:
R0 (x) = −0.08x + 800 USD per speaker system.
(c) (3 points) Compute R0 (5000).
Solution:
R0 (5000) = −0.08(5000) + 800 = 400 USD per speaker system.
(d) (3 points) In a complete English sentence, interpret your result in part (c).
Solution: At the production level of 5000 speaker systems, manufacturing the 5001st speaker
system will realise an additional revenue of approximately 400 U.S. Dollars.
Ogül Arslan/Eric Fu
Page 3 of 3
End of exam.