Download Exam 5 Solutions

Math 1513 - College Algebra
Exam #5 - 2016.11.16
Solutions
1. Let f (x) =
2x − 1
.
x+3
(a) State the domain and range of f (x).
The domain will be all numbers except where the denominator is zero, thus (−∞, −3) ∪ (−3, ∞).
The range is all real numbers except y = 2 – (−∞, 2) ∪ (2∞).
(b) Graph the function f (x).
(c) Algebraically compute f −1 (x).
We replace x with y and y with x and then solve for y:
x=
2y − 1
→ x(y + 3) = 2y − 1
y+3
→ xy + 3x = 2y − 1
→ 1 + 3x = 2y − xy
→ 1 + 3x = (2 − x)y
1 + 3x
→y=
2−x
(d) Graph the function f −1 (x).
1
2
(e) Verify that (f ◦ f −1 )(x) = x.
(f ◦ f
−1
)(x) = f
=
1 + 3x
2−x
2 1+3x
2−x − 1
1+3x
2−x
+3
2(1 + 3x) − 1(2 − x)
(1 + 3x) + 3(2 − x)
2 + 6x − 2 + x
=
1 + 3x + 6 − 3x
7x
=
7
=x
=
2. Sketch a graph of the function g(x) = 3−x+1 − 2.
We start with y = 3x , then shift to the right one unit: y = 3x−1 . Next, we reflect about the y−axis to get
y = 3−(x−1) . Finally, we shift the graph two units down to get g(x). The dashed blue curve below is the original
graph y = 3x , and the solid black line is our function g(x).
3. Determine the domain of h(x) = log42 −x2 (x − 1)(x + 1)3 .
We need to find all the values of the argument p(x) = −x2 (x − 1)(x + 1)3 such that p(x) > 0. The real line gets
broken into subintervals (−∞, −1) ∪ (−1, 0) ∪ (0, 1) ∪ (1, ∞) and we test each interval. The positive intervals are
(−1, 0) and (0, 1). Thus the domain of h(x) is (−1, 0) ∪ (0, 1).
4. Graph the function r(x) = log2 (2x − 1) + 1.
First we rewrite the function to get a better idea:
y = log2 (2x − 1) + 1
1
+1
= log2 2 x −
2
1
= log2 (2) + log2 x −
+1
2
1
+ 1 + log2 (2)
= log2 x −
2
3
So now we recognize that this is the graph of log2 (x) shifted to the right 12 units and then up 1 + log2 (2) units. The
dashed blue curve below is the original graph y = log2 (x), and the solid black line is our function r(x).
5. Use properties of logarithms to write the following expression as sums or differences of simple logarithmic terms.
s
!
2
4
3 (x + 1) (y − 2) (z + 1)
log3
(a + 5)7 (b + 1)2 (c − 4)4
We use the power rule first:
s
log3
3
(x + 1)2 (y − 2)4 (z + 1)
(a + 5)7 (b + 1)2 (c − 4)4
!
1
= log3
3
(x + 1)2 (y − 2)4 (z + 1)
(a + 5)7 (b + 1)2 (c − 4)4
Next we use the product and quotient rules:
1
(x + 1)2 (y − 2)4 (z + 1)
1
=
log3 ((x + 1)2 ) + log3 ((y − 2)4 ) + log3 (z + 1)
log3
7
2
4
3
(a + 5) (b + 1) (c − 4)
3
− log3 ((a + 5)7 ) − log3 ((b + 1)2 ) − log3 ((c − 4)4 )
Lastly, we use the power rule once more:
s
!
2
4
1
3 (x + 1) (y − 2) (z + 1)
= [2 log3 (x + 1) + 4 log3 (y − 2) + log3 (z + 1)
log3
7
2
4
(a + 5) (b + 1) (c − 4)
3
−7 log3 (a + 5) − 2 log3 (b + 1) − 4 log3 (c − 4)]
6. Solve for x in the following equation:
x+2
1
= 2 · 53x−1
3
We first take log of both sides (pick your favourite base):
x+2 !
1
log
= log 2 · 53x−1
3
Simplifying using the product and power rules gives
1
= log(2) + (3x − 1) log(5)
(x + 2) log
3
Next we multiply everything out:
1
1
x + 2 log
= log(2) + 3 log(5)x − log(5)
log
3
3
Everything with x’s in it gets moved to the left, and the rest to the right.
1
1
log
x − 3 log(5)x = log(2) − log(5) − 2 log
3
3
4
Now we factor x out of the left:
1
1
− 3 log(5) x = log(2) − log(5) − 2 log
log
3
3
And lastly, we divide by the expression multiplied by x on the left to finally solve for x.
log(2) − log(5) − 2 log 31
x=
log 31 − 3 log(5)
7. Solve for x in the following equation:
log(1 − x) + log(x) = log(x − 4)
This equation can have no solution, as the domains of each individual logarithmic function have an empty intersection. If we do try, we get the following:
log((1 − x)x) = log(x − 4)
Using 10 as a base for each side gives
(1 − x)x = x − 4
Solving for x gives x2 − 4 = 0 or x = ±2. Neither of those are valid solutions.
8. Solve for x in the following equation:
log3 (x − 2) = 1 − log3 (2x + 1)
We move the terms with x in them to the right
log3 (x − 2) + log3 (2x + 1) = 1
and then use the product property of logarithms
log3 ((x − 2)(2x + 1)) = 1
Using 3 as a base for each side
(x − 2)(2x + 1) = 3
Multiplying the left side out gives
2x2 − 3x − 2 = 3
Setting equal to zero:
2x2 − 3x − 5 = 0
Factoring gives
The solutions are x = −1 and x =
5
2.
(x + 1)(2x − 5) = 0.
Only the x = 25 solution is valid, as x = −1 is not in the domain of log3 (2x + 1).