MATH 2250 - 23228 MIDTERM EXAM 1 MODEL SOLUTION
SPRING 2012 - MOON
Write your answer neatly and show intermediate steps.
Calculator, computer and any electronic devices are not allowed.
(1) (0pt) Quick survey: Check one of numbers below.
In your mind, this class is...
Too easy
Moderate
1
2
3
4
5
6
Too difficult
7
(2) (5pts) For the function f (x) = x3 − 3x + 1, find the average rate of change over
the interval [1, 2].
Solution:
∆y
f (2) − f (1)
(23 − 3 · 2 + 1) − (13 − 3 · 1 + 1)
8−6+1−1+3−1
=
=
=
= 4.
∆x
2−1
2−1
1
• Writing the definition
f (2) − f (1)
: 2 pts.
2−1
• Plugging in correct values of function so obtain
: 3 pts.
• Getting correct answer 4: 5 pts.
Date: February 2, 2012.
1
(23 − 3 · 2 + 1) − (13 − 3 · 1 + 1)
2−1
MATH 2250 - 23228 MIDTERM EXAM 1 MODEL SOLUTION
2
(3) Find the limits.
x2 + 2x − 3
(a) (5pts) lim 2
x→1 x − 3x + 2
Solution:
x2 + 2x − 3
(x − 1)(x + 3)
x+3
1+3
lim 2
= lim
= lim
=
= −4.
x→1 x − 3x + 2
x→1 (x − 1)(x − 2)
x→1 x − 2
1−2
(x − 1)(x + 3)
: 3pts.
(x − 1)(x − 2)
• Computing the limit −4 correctly: 5pts.
• If you didn’t use correct mathematical language, some points may be
cut.
• Showing correct result of factorization
sin(4x)
x→0
x
Solution:
sin(4x)
sin(4x) 4
sin(4x)
sin(4x)
lim
= lim
· = lim
· 4 = 4 lim
= 4 · 1 = 4.
x→0
x→0
x→0
x
x
4 x→0 4x
4x
(b) (5pts) lim
sin(4x)
· 4: 3 pts.
x→0
4x
• Computing the limit 4 correctly: 5 pts.
• You have to use reasonable steps. You can’t get any score if you just write
your answer. If you didn’t use correct mathematical language, some points
may be cut.
• Showing the crucial step lim
MATH 2250 - 23228 MIDTERM EXAM 1 MODEL SOLUTION
3
(4) Find the limits.
3x − 2
x→∞ 2x + 7
Solution:
(a) (5pts) lim
2
2
3
−
lim
3−0
3
3x − 2
x→∞ x
x =
= lim
=
= .
lim
7
7
x→∞
x→∞ 2x + 7
2+0
2
2+
2 + lim
x→∞ x
x
2
3−
x : 3pts.
• Showing the crucial step lim
7
x→∞
2+
x
3
• Computing the limit correctly: 5pts.
2
• If you didn’t use correct mathematical language, some points may be
cut.
3−
(b) (5pts) lim+
x→1
x2
x−1
Solution:
1
x2
1
2
2
= lim+ x ·
= lim+ x
= 1 · ∞ = ∞.
lim
lim
x→1
x→1
x→1+ x − 1
x→1+ x − 1
x−1
1
2
• Showing the crucial step lim+ x
lim
: 3pts.
x→1
x→1+ x − 1
• Computing the limit ∞ correctly: 5pts.
• You have to use reasonable steps. You can’t get any score if you just
write your answer. If you didn’t use correct mathematical language,
some points may be cut.
MATH 2250 - 23228 MIDTERM EXAM 1 MODEL SOLUTION
(5) (10pts) For the function

1,





x,
g(x) = 1,



x,



1,
x ≤ −1
−1 < x < 0
x=0
0<x<1
x≥1
find points of discontinuity and explain your answer by using limits.
The graph of g(x)
Solution:
When x = −1,
lim g(x) = 1, lim + g(x) = −1.
x→−1−
x→−1
So lim g(x) does not exists and x = −1 is a discontinuity of g(x).
x→−1
At x = 0,
lim g(x) = lim+ g(x) = 0,
x→0−
x→0
so lim g(x) = 1. But g(0) = 1 6= 0 = lim g(x).
x→0
x→0
Therefore x = 0 is a discontinuity of g(x).
• Finding discontinuities: 2pts for each.
• Explaining correct reason: 2pts for each.
• Stating necessary values of limits and functions: 1pt for each.
4
MATH 2250 - 23228 MIDTERM EXAM 1 MODEL SOLUTION
5
df
(6) (a) (10pts) By using the definition of derivative, find the derivative
of
dx
√
f (x) = x + 2.
Solution:
√
√
df
f (x + h) − f (x)
x+h+2− x+h
= lim
= lim
h→0
h→0
dx
h
h
√
√
√
√
( x + h + 2 − x + h)( x + h + 2 + x + h)
√
√
= lim
h→0
h( x + h + 2 + x + h)
√
√
(x + h + 2) − (x + 2)
( x + h + 2)2 − ( x + 2)2
√
√
√
= lim √
= lim
h→0 h( x + h + 2 +
h→0 h( x + h + 2 +
x + 2)
x + 2)
h
1
√
√
= lim √
= lim √
h→0 h( x + h + 2 +
x + 2) h→0 x + h + 2 + x + 2
1
1
√
= √
= √
.
x+2+ x+2
2 x+2
f (x + h) − f (x)
: 2pts.
• Stating the definition of derivative lim
h→0
h
√
√
x+h+2− x+h
• lim
: 4pts.
h→0
h
√
√
√
√
( x + h + 2 − x + h)( x + h + 2 + x + h)
√
√
• Showing the step lim
:
h→0
h( x + h + 2 + x + h)
7pts.
1
correctly: 10pts.
• Computing the derivative √
2 x+2
• If you omit limit symbol, I cut 2pts more.
(b) (5pts) Find the equation of the tangent line of f (x) at x = 7.
Solution:
df 1
1
Slope of tangent line:
= √
= .
dx x=7 √
6
2 7+2
Tangential point: (7, f (7)) = (7, 7 + 2) = (7, 3).
1
Equation of tangent line: y = (x − 7) + 3.
6
• Calculating slope of tangent line: 2pts.
• Finding tangential point: 2pts.
• Calculating the equation of tangent line: 1pts.
MATH 2250 - 23228 MIDTERM EXAM 1 MODEL SOLUTION
6
(7) (10pts) For the function
(
sin x,
g(x) =
x,
x≤0
x>0
determine whether g(x) is differentiable or not at x = 0.
Solution:
To show the differentiability of g(x) at x = 0, we have to show the existence
g(0 + h) − g(0)
g(h) − g(0)
of g 0 (0) = lim
= lim
.
h→0
h→0
h
h
h−0
g(h) − g(0)
= lim+
= lim+ 1 = 1.
lim+
h→0
h→0
h→0
h
h
g(h) − g(0)
sin h − 0
sin h
lim−
= lim−
= lim−
= 1.
h→0
h→0
h→0
h
h
h
g(h) − g(0)
= 1.
Therefore, g 0 (0) = lim
h→0
h
So g is differentiable at x = 0 and g 0 (0) = 0.
• Computing one of one-sided limit: 5pts for each.