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

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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

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x

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2

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-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

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(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

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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

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t

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(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

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0

s 1

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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 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 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 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 0.

Solution: By parts (a), (b), and (c) above, we have considered all possible cases of a 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 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

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Figure 6: y = x2

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Figure 7: y = (x − 1)2

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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

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1

-1

1

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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

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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

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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

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(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

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Figure 13: y = log3 x

Figure 14: y = log3 (−x) y 6

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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

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1 x

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Figure 16: y = log x

Figure 17: y = log(x + 3)

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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.