Download y=|2x-2| y=|x-2| y=1 y=|x-2|

1
MATH 115.3 Midterm # 1 Solutions
y
1.
4
(a) (2%) Solve the equation |2x − 2| = |x − 2|.
4
Solution:
x = 0,
3
3
2
1
0
-1
x
0
1
2
3
y=|2x-2|
4
5
y=|x-2|
y
(b) (2%) Solve the inequality |x − 2| > 1.
Solution:
(−∞, 1) ∪ (3, ∞)
2
y=1
1
0
x
0
1
2
3
4
y=|x-2|
(c) (2%) Solve the inequality |x − 2| < −1.
Solution:
No Solution!
(d) (2%) Find a standard form equation of the line parallel to the line y = 3x − 1 passing
through the point (2,-1).
Solution: The slope of the required line is 3, so we first find the Point-Slope
equation y − (−1) = 3(x − 2) or y + 1 = 3x − 6. A standard form equation is then
−3x + y + 7 = 0.
(e) (2%) If a person’s temperature is 104◦ Fahrenheit, what is it in Celsius degrees?
Solution: C = 5 (104 − 32) = 5 (72) = 360 = 40
9
9
9
2
2.
3
3−2k+3
(a) (2%) Evaluate the exponential expression
.
34+k
3
3 3
3−2k+3
−2k+3−(4+k)
−3k−1
Solution:
=
3
=
3
= 33×(−3k−1) =
34+k
3−9k−3 .
(b) (2%) Which real number x satisfies log 1 3125 = x?
25
5
Solution: log 512 55 = ln 5−2 = 5 ln 5 = 5 = −2.5
ln 5
−2 ln 5
−2
2x+1
= 243 for x.
(c) (2%) Solve (27)
Solution: We have 33 2x+1 = 35 , or 33(2x+1) = 35 ,
so we must have 3(2x + 1) = 5, or 6x + 3 = 5, so 6x = 2, and thus
x=
1
.
3
(d) (2%) Solve ln x 5 − 2 ln x 2 = 1 for x.
Solution:
We have 5 ln x − 2(2 ln x) = 1, or 5 ln x − 4 ln x = 1, so ln x = 1.
Exponentiating, we have x = eln x = e1 , and thus
(e) (2%) Simplify the expression e−1.2 ln x .
−1.2
Solution: e−1.2 ln x = eln x
=
x −1.2 .
x = e.
3
3.
(a) (2%) Write the expression log5 (7x − 2) in terms of base e.
Solution:
ln(7x − 2)
ln 5
log5 (7x − 2) =
(b) (2%) Explain how the graph of y = −(x +1)2 +1 can be obtained from that of y = x 2 .
Solution:
Reflect about the x-axis, shift one unit left, and one unit up.
(c) (8%) Plot and label the numbers on the logarithmic scale provided below: 0.05,0.2,1,3,20,60,300,
and 800.
Solution:
0.05
1
2
0.2
3 4 5 67891
-2
2
1
3 4 5 67891
-1
10
10
0
10
3
2
20
3 4 5 67891
1
10
2
60
300
3 4 5 67891
2
10
2
800
3 4 5 6789
3
10
(d) (2%) A gas station’s tanks hold around 100,000 litres. Your car’s gas tank holds 50
litres. What is the approximate difference, expressed in terms of orders of magnitude?
Solution: 100, 000 = 2, 000 = 2.0 × 103 , so the two tanks differ in capacity by
50
about 3 orders of magnitude.
(e) (6%) When log y is graphed as a function of x on log-linear paper, a straight line
results. On Cartesian graph paper, the graph passes through the points (x1 , y1 ) =
(1, 400) and (x2 , y2 ) = (5, 80). Determine the functional relationship.
Solution:
The functional relationship must satisfy log y = mx + b, and we
also have
log 400 = m(1) + b and
log 80 = m(5) + b.
Subtracting the second equation from the first gives
1
400
log 5
log 400 − log 80
= − log
=−
.
log 400 − log 80 = −4m, so m =
−4
4
80
4
Using this value of m in the first equation, we get
log 5
log 5
b = log 400 − m = log 400 − −
= log 400 +
, and thus we have
4
4
log 5
log 5
log y = −
x + log 400 +
. Exponentiating, we have
4
4
14 − x4
log 5
log 5
log 5
log 5
y = 10− 4 x+log 400+ 4 = 10− 4 x ×10log 400 ×10 4 = 400× 10log 5 × 10log 5
=
1
x
400 × (5) 4 × (5)− 4 = 400 × 5
1−x
4
.
4
4. The following table is based on a functional relationship between x and y, which is either
an exponential or a power function.
x 0.10 0.50 1.00 1.50 2.00 2.50 3.00
y 7.46 5.67 4.00 2.83 2.00 1.41 1.00
(a) (10%) Use appropriate logarithmic transformations and graphs on both the graph
papers below to decide whether the table comes from a power function or an exponential function.
Solution:
Y=log y
Y=log y
9
8
7
6
5
9
8
7
6
5
4
4
3
3
2
2
1
0
1
2
3
4
5
6
7
8
9
10
x
1
2
3
4
1
5 6 7 8 91
2
3
4
5 6 7 8 9 10
(b) (10%) Find the functional relationship between x and y. Solution: Since the
graph on semilog paper appears to be closest to a straight line, we assume a functional relationship of the form log y = mx + b. Using the values (1, 4) and (2, 2),
we observe that the value of y is divided by 2 when x is increased by 1, so we have
y = 8 when x = 0, and the general functional relationship
t
1
y =8
.
2
X=log x
5
5.
1 3
(a) (3%) Differentiate h(t) =
t − 3t 2 + 12t + 3
3
1 3
1 3
2
t − 3t + 12t + 3
Solution: h (t) =
t − 3t 2 + 12t + 3 =
=
3
3
1 3 2 t
− 3t
+ (12t) + (3) =
3
1 3−1 − 3 t 2 + 12 t 1 + 0 =
3t
3
1 2
3t − 3 2t 2−1 + 12 (1)t 1−1 =
3
1 2
3t − 6t + 12 =
3
t 2 − 2t + 4
(b) (3%) Differentiate f (t) = t 3 e−2 + t
Solution: f (t) = t 3 e−2 + t = t 3 e−2 +(t) = e−2 t 3 + t 1 = e−2 3t 3−1 +
3 2
(1)t 1−1 = e−2 3t 2 + t 0 =
t +1
e2
1
(c) (3%) Differentiate y = √
,
3
t2
2
2 2
2 5
Since y = t − 3 , y = − t − 3 −1 = − t − 3
3
3
1
,
(d) (3%) Differentiate y = ln √ 3
t −t
Solution: We write y = − 1 ln(t 3 − t), so that y = − 1 3 1 (t 3 − t) =
2
2t −t
2
1 1
1
−
3t
− 3
(3t 2 − 1) =
2t −t
2(t 3 − t)
Solution:
(e) (3%) Differentiate f (t) = te2+t
Solution: f (t) = t e2+t + (t) e2+t = t(2 + t) e2+t + (1)e2+t = t(1) e2+t +
e2+t =
(t + 1)e2+t
et
t+1
t e (t + 1) − et (t + 1)
et (t + 1) − et (1)
f (t) =
=
=
(t + 1)2
(t + 1)2
(f) (3%) Differentiate f (t) =
Solution:
et
t
(t + 1)2
(g) (4%) Find an equation of the tangent line to y = 3x 2 − 1 at (1, 2).
Solution:
tangent line is
y = 6x = 6(1) when x = 1, so the point-slope equation of the
y − 2 = 6(x − 1)
6
6. Differentiate:
(a) (3%) h(t) = 2t
Solution:
t
h(t) = eln 2 = et ln 2 , so h (t) = (t ln 2) et ln 2 =
(ln 2)2t
(b) (3%) f (t) = t 2
Solution:
f (t) = 2t 2−1 =
2t
2
(c) (3%) y = 2 ,
Solution:
(d) (3%) y = t t ,
y = 0
t
Solution: y = e
= et ln t , so y = (t ln t) et ln t = (t(ln t) + (t) ln t) et ln t =
1
t
+ (1) ln t t t = (1 + ln t)t t
t
ln t
2
(e) (3%) y = e3t ,
Solution:
2
y = 3t 2 e3t =
6te3t
2
(f) (3%) f (t) = ln(3t 2 + 1)
Solution:
f (t) =
2
1
3t + 1 =
+1
3t 2
6t
+1
3t 2