Vanier College
Department of Mathematics
201-015-50
Sec V Mathematics
Test 2 - Solutions
Winter 2011
/10
1. (a) Find the domain of the following function.
r
(3 − x)(16 − x2 )
f (x) =
x2 − 12x + 35
(b) Find the horizontal and vertical asymptotes of the following rational function.
f (x) =
/10
−18x3 + 15x2 + 2x
3x3 + 2x2 − x
2. Let
f (x) =
1 − 2x
,
2−x
g(x) = 3 − 2x,
h(x) = 1 − ln x,
and
k(x) = e1−x .
Find
(a) (f ◦ g)(x)
(b) (h ◦ k)(x)
(c) g −1 (x)
(d) f −1 (x)
/10
3. Write the following expression as a single logarithm.
√
(a) 21 log3 3 x + log3 (9x2 ) − log3 9
√
(b) 2 log2 4 + 3 log2 2 − 2 log2 54
/15
4. Consider the following function.
f (x) = −2 · 3x + 6
(a) Draw the graph of f .
(b) Find Dom(f )
(c) Find R(f )
(d) Find the zeros of f
(e) Find the y-intercept
(f) Analyze the variation of f
(g) Find the extremums of f
(h) Analyze the sign of f
(i) Find f −1 (x)
(e) h−1 (x)
/15
5. Consider the following function.
1
f (x) = log2 − (x − 4)
2
(a) Draw the graph of f .
(b) Find Dom(f )
(c) Find R(f )
(d) Find the zeros of f
(e) Find the y-intercept
(f) Analyze the variation of f
(g) Find the extremums of f
(h) Analyze the sign of f
(i) Find f −1 (x)
/10
6. Solve the following equations.
(a) log4 (x + 3) + log4 (2 − x) = 1
(b) 32x − 3x − 12 = 0
/10
7. A car loses a certain percentage of its value every year. After 4 years, it is worth 13 040.40$.
After 8 years, it is worth 6 940.53$.
(a) What percentage of its value does the car lose every year?
(b) What is the price of such a car when it’s new?
(c) After how many years will the car be worth 4 897.24$?
/10
8. 5 000$ are invested in a fund that yeilds 6% interest annually.
(a) How much will the fund be worth in 24 years if the interest is compounded annually?
(b) How much will the fund be worth in 24 years if the interest is compounded monthly?
(c) How much will the fund be worth in 24 years if the interest is compounded daily?
(d) In how much time will the fund be worth in 103 484.43$ if the interest is compounded
yearly?
/5
9. If logb x = a , then log√b
(a)
/5
3a
2
(b)
√
3
x2 equals
3a
4
(c)
4a
3
4
(d) a 3
10. If 4x = 3, find the value of 24x
(a) 24
(b) 18
(c) 8
(d) 9
(d)
SOLUTIONS
1 − 2y
2−y
x(2 − y)1 − 2y
1. (a)
r
f (x) =
x=
(3 − x)(16 − x2 )
x2 − 12x + 35
2y − xy = 1 − 2x
We find Dom(f ) by letting
y(2 − x) = 1 − 2x
1 − 2x
y=
2−x
1
− 2x
→ f −1 (x) =
2−x
(3 − x)(16 − x2 )
≥ 0,
x2 − 12x + 35
which is equivalent to:
(3 − x)(4 − x)(4 + x)
≥ 0.
(x − 7)(x − 5)
(e)
After building our sign table, we get:
x = 1 − ln y
Dom(f ) = [−4, 3] ∪ [4, 5[ ∪ ]7, ∞[
ln y = 1 − x
y = e1−x
(b)
→ h−1 (x) = e1−x
−18x3 + 15x2 + 2x
f (x) =
3x3 + 2x2 − x
Horizontal Asymptote:
3
3. (a)
2
−18x + 15x + 2x
= −6
x→+∞
3x3 + 2x2 − x
→ y = −6
21 log3
lim
x7 (9x2 )
9
= log3 x9
= log3
3x3 +2x2 −x = 0 → x(3x−1)(x+1) = 0
1
, x = −1
3
2. (a)
x + log3 (9x2 ) − log3 9
= log3 x7 + log3 (9x2 ) − log3 9
Vertical Asymptote:
→ x = 0, x =
√
3
(b)
2 log2 4 + 3 log2 2 − 2 log2
1 − 2(3 − 2x)
(f ◦ g)(x) =
2 − (3 − 2x)
4x − 5
=
2x − 1
(b)
(h ◦ k)(x) = 1 − ln e1−x
= 1 − (1 − x)
=x
(c)
x = 3 − 2y
2y − x + 3
1
y =− x+
2
1
−1
→ g (x) = − x +
2
3
2
3
2
√
54
= log2 42 + log2 23 − log2 54
42 · 23
54
64
= log2
27
= log2
4.
5.
f (x) = −2 · 3x + 6
f (x) = log2
1
− (x − 4)
2
(a) .
(a) .
(b) Dom(f ) = R
(c) R(f ) =] − ∞, 6[
(b) Dom(f ) =] − ∞, 4[
(d) zero: 1
(c) R(f ) = R
(e) y-intercept: 4
(d) zero: 2
(f) Variation of f :
f % if x ∈ ∅
f & if x ∈ R
(e) y-intercept: 1
(g) Extremums of f : None
(h) Sign of f
f ≥ 0 if x ∈ ] − ∞, 1]
f ≤ 0 if x ∈ [1, ∞[
(f) Variation of f :
f % if x ∈ ∅
f & if x ∈ ] − ∞, 4[
(g) Extremums of f : None
(h) Sign of f
f ≥ 0 if x ∈ ] − ∞, 2]
f ≤ 0 if x ∈ [2, 4[
(i) f −1 (x) =?
x = −2 · 3y + 6
→ x − 6 = −2 · 3y
1
→ − (x − 6) = 3y
2
1
→ y = log3 − (x − 6)
2
(i) f −1 (x) =?
1
x = log2 − (y − 4)
2
1
→ − (y − 4) = 2x
2
→ y − 4 = −2 · 2x
→ y = −2 · 2x + 4
1
→ f −1 (x) = log3 − (x − 6)
2
→ f −1 (x) = −2 · 2x + 4
6. (a)
9.
log4 (x + 3) + log4 (2 − x) = 1
log√
Restrictions:
We must have
√
3
b
x2
=
=
x + 3 > 0 → x > −3
=
and
=
2−x>0→x<2
√
3
log x2
√
log b
2
3 log x
1
2 log b
4
logb x
3
4a
3
log4 (x + 3) + log4 (2 − x) = 1
Answer: C
→ log4 (x + 3)(2 − x) = 1
10.
→ (x + 3)(2 − x) = 4
2
→ −x − x + 6 = 4
24x = (22 )2x
2
→x +x−2=0
= 42x
→ (x + 2)(x − 1) = 0
= (4x )2
= 32
=9
Therefore, x = −2 or x = 1, which
are both acceptable answers.
(b)
32x − 3x − 12 = 0
Answer: D
Let y = 3x . The equation becomes:
y 2 − y − 12 = 0 → (y − 4)(y + 3) = 0
Therefore,
−3.
Then y = 4 or y=
x
3 = 4 → x = log3 4
7. (a) We must solve the following system:
13040.40 = a · c4
6940.53 = a · c8 .
We get c = 0.854, which means that
the car loses around 14.6% of its value
every year.
(b) To find the initial price of the car, we
must solve:
13040.40 = a · (0.854)4 .
We get: a = 24501.30$
(c) We solve 4897.24 = 24501.30·(0.854)x ,
and get: x = 10.2
24
8. (a) y = 5000 · (1 + 0.06) = 20244.67$
24·12
0.06
= 21027.89$
(b) y = 5000· 1 +
12
24·365
0.06
(c) y = 5000· 1 +
= 21100.98$
365
x
(d) 103484.43 = 5000 · (1 + 0.06)
→ x = 52 years.