Download Math 115 Spring 11 Written Homework 12 Solutions

Math 115 Spring 11
Written Homework 12 Solutions
1. Consider f (x) := (x − 2)2 (x + 2)2 .
(a) What is the domain of f ? Justify your answer.
Solution: Since f is a polynomial, we know that the domain of f is all of R.
(b) What is the range of f ?
Solution: Note that f is the product of the two terms (x − 2)2 and (x + 2)2 and both of
these terms are greater than or equal to 0 for all values of x, the range of f is [0, ∞).
(c) Sketch a graph of f using long-run behavior and intercepts.
• Long-Run Behavior: Since f is a polynomial, the long-run behavior is dictated by the
highest order term of the polynomial. For f , the highest order term is (x2 )(x2 ) = x4 .
Since the highest degree of f is even, we also know that
lim f (x) = lim f (x) = lim x4 = ∞.
x→−∞
x→+∞
x→+∞
• Intercepts
– x-intercepts: Since f is already in factored form, we see the x-intercepts to be
(−2, 0) and (2, 0).
– The y-intercept occurs at (0, f (0)) = (0, 16).
Combining these conditions, we get a frame for our graph: Using the domain, range and the
y
20
15
10
5
-4
x
-2
0
2
4
knowledge that f is a continuous function, we get something that looks like:
y
20
15
10
5
-4
-2
x
0
2
Figure 1: y = (x − 2)2 (x + 2)2
4
(d) Is f a one-to-one function? If not, describe a domain on which f is one-to-one.
Solution: Note that the graph spectacularly fails the horizontal line test. Given any line
y = k that intersects the graph y = f (x), the horizontal line will intersect the curve in at
least 2 points (if not three or four). Hence, f (x) is not a one-to-one function. However, the
graph can be broken into pieces on which f is one-to-one. The biggest such intervals are
(−∞, −2], [−2, 0], [0, 2] and [2, ∞).
(e) Does your domain in (d) cover the entire range stated in (b)? If not, find another domain on which f is one-to-one such that the image of the restricted domain is the entire range.
Solution: Recall that the range of f being [0, ∞) is represented graphically by the y values
of the points on the graph. The intervals x ∈ (−∞, −2] and x ∈ [2, ∞) both cover the entire
range of f .
2. For the following functions f and g, verify that they are inverses by showing f (g(x)) := x
and that g(f (x)) := x.
(a) f (x) := x3 + 3 and g(x) := (x − 3)1/3
Solution:
f (g(x)) := ((x − 3)1/3 )3 + 3
= (x − 3) + 3
= x−3+3
= x
g(f (x)) := ((x3 + 3) − 3)1/3
= (x3 + 3 − 3)1/3
= (x3 )1/3
= x
Since f (g(x)) := x and g(f (x)) := x, the functions f and g are inverses.
(b) f (x) :=
x+1
x+1
and g(x) :=
x−1
x−1
Solution: Note that here f and g are the same function. Hence f (g(x)) = g(f (x)). In other
words, we are showing that the inverse of the function f is f itself.
x+1
+1
x−1
f (g(x) := x+1 −1
x−1
=
=
x+1
+ x−1
x−1
x−1
x+1
x−1
−
x−1
x−1
(x+1)+(x−1)
x−1
(x+1)−(x−1)
x−1
(x + 1) + (x − 1)
(x + 1) − (x − 1)
x+1+x−1
=
x+1−x+1
2x
=
2
= x
=
Hence, the functions f and g are inverses.
3. Find a domain on which f is one-to-one and a formula for the inverse of f restricted to
this domain. Sketch the graphs of f and f −1 .
1
(a) f (x) :=
x+1
Solution: The domain of f is (−∞, −1) ∪ (−1, ∞). We know what the graph of the function
1
looks like. It is the graph of translated to the left by 1.
x
y
4
2
-4
x
-2
2
4
-2
-4
Figure 2: y = f (x)
Since the graph of f passes the horizontal line test, we know that f is a one-to-one
function on its domain. Hence, it is invertible.
To find f −1 , we reverse the roles of x and y in the equation y = f (x) and solve for y.
1
y+1
x(y + 1) = 1
1
y+1=
x
1
y = −1
x
1
Therefore, we define the inverse of f to be f −1 (x) = − 1.
x
There are multiple ways to graph this function. The easiest being to realize that this is
x =
the graph y = 1/x shifted down 1. Here, I am going to demonstrate that the graphs y = f (x)
(the blue curve) and y = f −1 (x) (the green curve) are symmetric about the line y = x (the
red line).
y
4
2
-4
x
-2
2
4
-2
-4
(b) g(t) := t3
Solution: The domain of g is (−∞, ∞) since g is a polynomial. We know what the graph
of the is function looks like.
y
4
2
-4
t
-2
2
4
-2
-4
Figure 3: y = t3
Since the graph of g passes the horizontal line test, we know that g is a one-to-one
function on its domain. Hence, it is invertible.
To find g −1 , we reverse the roles of t and y in the equation y = g(t) and solve for y.
t = y3
√
3
t = y
√
3
t
y =
Therefore, we define the inverse of g to be g −1 (t) =
√
3
t.
Again, to graph the inverse of g, I am going to demonstrate that the graphs y = g(t)
(the blue curve) and y = g −1 (t) (the green curve) are symmetric about the line y = t (the
red line).
y
4
2
-4
t
-2
2
4
-2
-4
(c) h(s) :=
s2
1
+1
Solution: Note that s2 + 1 6= 0 for any value of s. (In fact, s2 + 1 ≥ 1 for all values of s.)
1
Therefore, the domain of h(s) := 2
is all of R.
s +1
To begin sketching this, again we use intercepts and long-run behavior.
• Long-Run Behavior Since f is a rational function, the long-run behavior is the same
at the ends ∞ and −∞. Therefore
lim h(s) = lim h(s) = lim
s→−∞
s→+∞
s→+∞ s2
1
= 0.
+1
• Intercepts
– x-intercepts Since h is already in factored form, we see that the numerator 1 never
equals 0. Hence, there are no x-intercepts.
– The y-intercept occurs at (0, h(0)) = (0, 1). (Note that, since the denominator is
smallest at s = 0, the point must be the location of the maximum value of h on
the graph.)
Combining these conditions, we get a frame for our graph:
y
2.0
1.5
1.0
0.5
-4
s
-2
2
4
-0.5
-1.0
Using the domain and the knowledge that f is a continuous function, we get something
that looks like:
y
2.0
1.5
1.0
0.5
-4
s
-2
2
4
-0.5
-1.0
Figure 4: y =
s2
1
+1
Note y = h(s) does not pass the horizontal line test. However, like the y = x2 example
in lecture or in problem 5e of this homework, we can restrict the domain of h to pieces on
which h is one-to-one. Here, the largest such intervals are (−∞, 0] and [0, ∞). I will choose
to work with the interval [0, ∞). On this interval, the one-to-one graph of y = h(s) looks
like
y
2.0
1.5
1.0
0.5
0.0
s
1
2
3
4
5
-0.5
-1.0
Figure 5: y =
s2
1
on [0, ∞)
+1
To find h−1 , we reverse the roles of s and y in the equation y = h(s) and solve for y.
s =
y2
1
+1
s(y 2 + 1) = 1
1
y2 + 1 =
s
1
y2 =
−1
s
r
1
− 1 (**)
y =
s
(**): Note that since we are interested in our inverse function returning us to the domain
s ∈ [0, ∞), we choose to use the positive square root in manipulating
this last line.
r
1
Therefore, we define the inverse of h to be h−1 (s) =
− 1.
s
Again, to graph the inverse of h, I am going to demonstrate that the graphs y = h(s)
(the blue curve) and y = h−1 (s) (the green curve) are symmetric about the line y = s (the
red line).
y
5
4
3
2
1
0
s
1
-1
2
3
4
5
4. A function g(t) is called a strictly increasing function if for every a and b in g’s domain
where a < b, then g(a) < g(b).
(a) Show that the function h(x) := x2 is NOT a strictly increasing function.
Solution: We just need one counter-example to show that h(x) is not strictly increasing.
Let a = −2 and let b = 1. Clearly a < b. But h(−2) = 4 > h(1) = 1. Hence, h is not strictly
increasing.
(b) Assume that g(t) is a strictly increasing function. Using the definition of one-to-one,
show that the strictly increasing function g(t) must by one-to-one.
Solution: Since g(t) is strictly increasing, g(a) < g(b) when a < b. Recall that one of the
equivalent definitions of a one-to-one function is that given g(t1 ) = g(t2 ), the two values t1
and t2 are equal. Here, since g is strictly increasing , if g(a) = g(b) then a must equal b.
Hence, g(t) is a one-to-one function.
(c) Explain how the graph of a strictly increasing function g(t) behaves. Why are you
guaranteed that a strictly increasing function passes the horizontal line test?
Solution: By definition, as t increases g(t) increases. Hence, on the graph (t, g(t)) as t
increases in value the height of the points go up. The graph will pass the horizontal line test
because it is one-to-one.
(d) Use the definition of a strictly increasing function to show that an equivalent condition
for strictly increasing is that for every a and b in g’s domain where a < b, then g(b)−g(a) > 0.
Solution: This is straight-forward algebra. (The point of this problem is that we will need
this fact below in problem 5.) Since g is strictly increasing, g(b) > g(a) when b > a. Hence,
subtracting the value of g(a) from both sides of the first inequality, g(b) − g(a) > 0.
5b0 . Let a and b be any real numbers. Consider the number b2 + ab + a2 .
(a) Show that if 0 < a < b that b2 + ab + a2 > 0. (Easy.)
Solution: Since a and b are both positive, we know that b2 , a2 , and ab are each positive.
Thus, b2 + ab + a2 is positive.
(b) Show that if a < b < 0 that b2 + ab + a2 > 0. (Easy.)
Solution: Since a and b are both negative, we know that b2 , a2 , and ab are each positive.
Thus, b2 + ab + a2 is positive.
(c) Show that if a < 0 < b that b2 + ab + a2 > 0. (Harder: Try completing the square.)
Solution: Here, we will complete the square on the term b2 + ab. We are viewing b as the
variable and letting a act as the coefficient of the linear term.
a 2 a 2
b2 + ab + a2 = b2 + ab +
−
+ a2
2
2
a 2 a2
+ a2
−
= b+
2
4
a 2 3a2
= b+
+
2
4
As argued above, every term is the final expression is squared and thus positive.
(d) Explain why, for any a < b that b2 + ab + a2 > 0.
Solution: By parts (a), (b), and (c) above, we have considered all possible cases of a < b
and in each case, we found b2 + ab + a2 > 0.
5. Let f (x) := x3 + x + 1.
(a) State the domain and range of f .
Solution: The domain of f is all real numbers since f is a polynomial. To determine the
range of f , we need to understand the long-run behavior of f . Note that
lim f (x) = lim x3 = ∞ and
x→∞
x→∞
lim f (x) = lim x3 = −∞.
x→−∞
x→−∞
Since f is a continuous function, the range of f must be all real numbers.
(b) Show that the function f (x) := x3 + x + 1 is a strictly increasing function. (Hint:
Explain why a < b implies that b − a > 0 for all real numbers a and b. Now use this to show
that f (b) − f (a) > 0 is a true statement. Your going to need to factor a cubic polynomial.)
Solution: As argued in problem 4d, if a < b, it is trivial that b − a > 0. Here, we need to
show that f (b) − f (a) > 0. We start by looking at f (b) − f (a).
f (b) − f (a) = b3 + b + 1 − (a3 + a + 1)
= b3 + b + 1 − a3 − a − 1
= b 3 − a3 + b − a
We know that b − a > 0. We need to show that b3 − a3 is greater than or equal to zero
always. As suggested in the hint, we hope that we can factor the term b − a from this cubic
polynomial. Using polynomial long-division,
b3 − a3 = (b − a)(b2 + ab + a2 ).
Then
f (b) − f (a) = b3 − a3 + b − a = (b − a)(b2 + ab + a2 ) + (b − a).
Since b − a > 0 and we proved in 5b0 that b2 + ab + a2 > 0 whenever b > a,
f (b) − f (a) > 0 always.
(c) Explain why part (b) proves that f is a one-to-one function.
Solution: As shown in problem 4 part b, if f is a strictly increasing function, we know that
it is a one-to-one function.
(c) Explain why f −1 exists (but do not attempt to find it).
Solution: By theorem, we know that every one-to-one function is an invertible function.
(d) Find f −1 (1).
Solution: We want the value of a in the domain of f (x) such that a = f −1 (1). By the
definition of inverses, this is equivalent to the value of a where f (a) = 1. That is, we want
a such that
a3 + a + 1 = 1.
By inspection, we realize that a must be 0.
Hence f −1 (1) = 0.
(e) Find f −1 (3).
Solution: Again, we want the value of a in the domain of f (x) such that a = f −1 (3). By
the definition of inverses, this is equivalent to the value of a where f (a) = 3. That is, we
want a such that
a3 + a + 1 = 3.
By inspection, we realize that a = 1.
Hence f −1 (3) = 1.
6. Let f (x) := x2 − 2x. Determine a domain on which f −1 exists and find a formula for f −1
on this domain.
Solution: The domain of f is all of R, since f is a polynomial function. To determine an
interval on which f is one-to-one, we need to look at the graph of f .
Note that y = x2 − 2x is the graph of a parabola. One way to graph this is to complete
the square and use transformations. Note that
y = x2 − 2x = x2 − 2x + 1 − 1 = (x − 1)2 − 1.
Now start with the graph of the basic power function y = x2 . Then graph y = (x − 1)2 ,
the horizontal shift of the basic graph one unit to the right. Lastly, apply the vertical shift
y = (x − 1)2 − 1.
y
-4
y
3
3
2
2
1
1
-2
2
-1
Figure 6: y = x2
4
x
-4
-2
2
-1
Figure 7: y = (x − 1)2
4
x
y
3
2
1
-4
-2
2
4
x
-1
Figure 8: y = (x − 1)2 − 1 = x2 − 2x
(Remark: We could also have used long-run behavior and intercepts to sketch a rough
graph. Without remembering that the graph of y = x2 − x is a parabola, it is hard to find
the exact point of the graph where the value of f is a minimum. We need this determine
the maximum intervals on which f is one-to-one.)
As with the example h(x) := x2 in lecture, we see that the graph of y = x2 − x passes
the horizontal line test when the domain of f is restricted to (−∞, 1] or [1, ∞).
I will choose to find an inverse on the interval [1, ∞). On this interval, the one-to-one
graph of y = f (x) looks like
y
4
3
2
1
-1
1
2
3
4
x
-1
Figure 9: y = x2 − 2x on [1, ∞)
To find f −1 , we reverse the roles of x and y in the equation y = f (x) and solve for y.
x = y 2 − 2y
y 2 − 2y − x = 0
To solve for y, we need to apply the quadratic formula. Here, the coefficients are A = 1,
B = −2 and C = −x. Then
p
√
√
−(−2) ± (−2)2 − 4(1)(−x)
2 ± 4 + 4x
=
= 1 ± 1 + x.
y=
2(1)
2
(We could also have solved for this by completing the square.) Note that since we are
interested in our inverse function returning us to the domain x ∈ [1, ∞), we need to choose
to use the positive square root in the solution for y.
Therefore, we define the inverse of f to be f −1 (x) = 1 +
√
1 + x.
Again, to graph the inverse of f , I am going to demonstrate that the graphs y = f (x)
(the blue curve) and y = f −1 (x) (the green curve) are symmetric about the line y = x (the
red line).
y
4
3
2
1
-1
1
-1
2
3
4
x
7. Determine the exact value of the following expressions:
(a) log(4) + 2 log(5)
Solution: Use the power rule and the product rule for logarithms:
log(4) + 2 log(5) = log(4) + log(52 ) = log[(4)(52 )] = log(100) = 2
(b)
log4 49
log4 7
Solution:
log4 49
log4 72
2 log4 7
=
=
=2
log4 7
log4 7
log4 7
(c) ln
1
e2
Solution:
ln
1
e2
= ln(e−2 ) = −2 ln e = −2
8. Fully expand this expression: logπ
a4 b 5
.
cd3
Solution: Use the quotient rule, followed by the product rule, followed by the power rule:
logπ
a4 b 5
cd3
= logπ (a4 b5 ) − logπ (cd3 )
= logπ a4 + logπ b5 − (logπ c + logπ d3 )
= 4 logπ a + 5 logπ b − logπ c − 3 logπ d
9. Using a proper sequence of transformations, graph the following functions. For each
graph, clearly label any asymptotes and at least one point on the graph.
(a) f (x) = − ln(x − 2)
Solution: We will begin with the graph of y = ln x. Then reflect that graph of the x-axis
and shift it horizontally to the right 2.
y
y
3
3
2
2
1
1
-1
1
2
3
4
5
x
-1
1
-1
-1
-2
-2
-3
-3
2
Figure 11: y = − ln x
Figure 10: y = ln x
y
3
2
1
-1
3
1
2
3
4
-1
-2
-3
Figure 12: y = − ln(x − 2)
5
x
4
5
x
(b) g(x) = 4 + log3 (−x)
Solution: We will begin with the graph of y = log3 x. Then reflect that graph over the
y-axis and shift it vertically up 4.
y
-4
y
6
6
4
4
2
2
-2
2
4
x
-4
-2
2
-2
-2
Figure 13: y = log3 x
Figure 14: y = log3 (−x)
y
6
4
2
-4
-2
2
-2
Figure 15: y = 4 + log3 (−x)
4
x
4
x
(c) h(x) = log(x + 3)
Solution: We will begin with the graph of y = log x. Then shift it horizontally to the left
3.
y
-4
y
3
3
2
2
1
1
x
-2
2
4
-4
x
-2
2
-1
-1
-2
-2
-3
-3
Figure 16: y = log x
Figure 17: y = log(x + 3)
4
10. Determine the domain of each of the following functions:
(a) f (x) := ln(3x + 2)
Solution:
3x + 2 > 0
3x > −2
2
x > −
3
The domain is D :
(b) g(x) := log4
2
− ,∞ .
3
√
5 − 3x
Solution:
√
5 − 3x > 0
5 − 3x > 0
−3x > −5
5
x <
3
The domain is D :
5
−∞,
.
3
11. Explain why it does not make sense to ask for the value of lim (ln x).
x→−∞
Solution: The domain of y = ln x is all real numbers greater than 0: x > 0. Thus, negative
values are not in the domain so it makes no sense to ask about the behavior as x goes to −∞.
12. Solve each of the following equations:
(a) e3x = 2507
Solution:
e3x = 2507
ln e3x = ln(2507)
(3x)(ln e) = ln(2507)
(3x)(1) = ln(2507)
3x = ln(2507)
ln 2507
x =
3
(b) 7x = 42x−1
Solution: We can use whatever base value we want to solve this equation. Here we will use
log base 10:
7x = 42x−1
log 7x = log 42x−1
(x) log 7 = (2x − 1) log 4
x log 7 = (2x) log 4 − (1) log 4
x log 7 − 2x log 4 = − log 4
x(log 7 − 2 log 4) = − log 4
− log 4
x =
log 7 − 2 log 4
(c) log3 (x − 4) + log3 7 = 2
Solution:
log3 (x − 4) + log3 7 = 2
log3 (7(x − 4)) = 2
32 = 7(x − 4)
9 = 7x − 28
37 = 7x
37
= x
7
Looking back at the original equation, the domain of the function y = log3 (x − 4) is x > 4.
This value falls in that domain and thus is a solution of the equation.
(d) log2 (x − 1) − log2 (x + 1) = 3
Solution:
log2 (x − 1) − log2 (x + 1) = 3
x−1
log2
= 3
x+1
x−1
23 =
x+1
x−1
8 =
x+1
8(x + 1) = x − 1
8x + 8 = x − 1
7x = −9
9
x = −
7
Looking back at the original equation, the domain of log2 (x − 1) is x > 1 and the domain of
log2 (x + 1) is x > −1 so overall a solution of this equation must fall in the interval (1, ∞).
Since x = − 79 does not fall in that interval, this equation has no solution.
(e) ln(x − 1) + ln 6 = ln(3x)
ln(x − 1) + ln 6 = ln(3x)
ln(6(x − 1)) = ln(3x)
6(x − 1) = 3x
6x − 6 = 3x
3x = 6
x = 2
Looking back at the original equation, the domain of ln(x − 1) is x > 1 and the domain of
ln(3x) is x > 0 so overall a solution of this equation must fall in the interval (1, ∞). Thus,
x = 2 is a solution of the equation.
(f) log(x + 14) − log x = log(x + 6)
log(x + 14) − log x
x + 14
log
x
x + 14
x
x + 14 =
= log(x + 6)
= log(x + 6)
= x+6
= x(x + 6)
x + 14 = x2 + 6x
x2 + 5x − 14 = 0
(x + 7)(x − 2) = 0
The two solutions to this quadratic are x = −7 and x = 2. But x = −7 is not in the domain
of log(x + 14) (or log(x + 6)). Thus, the only solution to this equation is x = 2.