## Engage NY Eureka Math Precalculus Module 1 Lesson 2 Answer Key

### Eureka Math Precalculus Module 1 Lesson 2 Exercise Answer Key

Exercise 1.

Let f(x) = sin(x). Does f(2x) = 2f(x) for all values of x? Is it true for any values of x? Show work to justify your answer.

Answer:

No. If x = \(\frac{\pi}{2}\), sin (2(\(\frac{\pi}{2}\))) = sin(Ï€) = 0, but 2sin(\(\frac{\pi}{2}\)) = 2(1) = 2, so the statement does not hold for every value of x. It is true anytime sin(x) = 0., so for x = 0,

x = Â±Ï€, x = Â±2Ï€.

Exercise 2.

Let f(x) = log(x). Find a value for a such that (2a) = 2f(a). Is there one? Show work to justify your answer.

Answer:

log(2a) = 21og(a)

log(2a) = log(a^{2})

2a = a^{2}

a^{2} – 2a = 0

a(a – 2) = 0

Thus, a = 2 or a = 0. Because 0 is not in the domain of the logarithmic function, the only solution is a = 2.

Exercise 3.

Let f(x) = 10^{x}. Show that f(a + b) = f(a) + f(b) is true for a = b = log(2) and that it is not true for a = b = 2.

Answer:

For a = b = log(2)

f(a + b) = 1o^{(log(2)+log(2))} = 10^{2log(2)} = 10^{log(22)} = 2^{2} = 4

f(a) + f(b) = 10^{log2} + 10^{log2} = 2 + 2 = 4

Therefore, f(a + b) = f(a) + f(b).

For a = b = 2

f(a + b) = 10^{2+2} = 10^{4} = 10000

f(a) + f(b) = 10^{2} + 10^{2} = 100 + 100 = 200

Therefore, f(a + b) â‰ f(a) + f(b).

Exercise 4.

Let f(x) = \(\frac{1}{x}\). Are there any real numbers a and b so that f(a + b) = f(a) + f(b)? Explain.

Answer:

Neither a nor b can equal zero since they are in the denominator of the rational expressions.

\(\frac{1}{a+b}\) = \(\frac{1}{a}\) + \(\frac{1}{b}\)

\(\frac{1}{a+b}\)ab(a + b) = \(\frac{1}{a}\)ab(a + b) + \(\frac{1}{b}\)ab(a + b)

ab = a(a + b) + b(a + b)

ab = a^{2} + ab + ab + b^{2}

ab = a^{2} + 2ab + b^{2}

ab = (a + b)^{2}

This means that ab must be a positive number. Simplifying further, we get 0 = a^{2} + ab + b^{2}.

The sum of three positive numbers will never equal zero, so there are no real solutions for a and b.

Exercise 5.

What do your findings from these exercises illustrate about the linearity of these functions? Explain.

Answer:

Answers will vary but should address that in each case, the function is not a linear transformation because it does not hold to the conditions f(a + b) = f(a) + f(b) and f(cx) = c(f(x)) for all real-numbered inputs.

### Eureka Math Precalculus Module 1 Lesson 2 Problem Set Answer Key

Assign students some or all of the functions to investigate. Problems 1-4 are all trigonometric functions, Problem 5 is a rational function, and Problems 6 and 7 are logarithmic functions. These can be divided up. Problem 8 sets up Lesson 3 but is quite challenging.

Examine the equations given in Problems 1-4, and show that the functions f(x) = cos(x) and g(x) = tan(x) are not linear transformations by demonstrating that they do not satisfy the conditions indicated for all real numbers. Then, find values of x and/or y for which the statement holds true.

Question 1.

cos(x + y) = cos(x) + cos(y)

Answer:

Answers that prove the statement false will vary but could include x = 0 and y = 0.

This statement is true when x = 1. 9455, or 111. 47Â°, and y = 1. 9455, or 111. 47Â°. This will be difficult for students to find without technology.

Question 2.

cos(2x) = 2cos(x)

Answer:

Answers that prove the statement false will vary but could include x = 0 or x = \(\frac{\pi}{2}\)

This statement is true when x = 1. 9455, or 111. 47Â°. This will be difficult for students to find without technology.

Answer:

Question 3.

tan(x + y) = tan(x) + tan(y)

Answer:

Answers that prove the statement false will vary but could include x = \(\frac{\pi}{4}\) and y = \(\frac{\pi}{4}\).

This statement is true when x = 0 and y = 0.

Question 4.

tan(2x) = 2tan(x)

Answer:

Answers that prove the statement false will vary but could include x = \(\frac{\pi}{4}\) and y = \(\frac{\pi}{4}\).

This statement is true when x = 0 and y = 0.

Question 5.

Let f(x) = \(\frac{1}{x^{2}}\). Are there any real numbers a and b so that f(a + b) = f(a) + f(b)? Explain.

Answer:

Neither a nor b can equal zero since they are in the denominator of the fractions.

If f(a + b) = f(a) + f(b), then \(\frac{1}{(a+b)^{2}}\) = \(\frac{1}{a^{2}}\) + \(\frac{1}{b^{2}}\).

Multiplying each term by a^{2}b^{2}(a + b)^{2}, we get

a^{2}b^{2}(a + b)^{2}\(\frac{1}{(a+b)^{2}}\)= a^{2}b^{2}(a + b)^{2}\(\frac{1}{a^{2}}\)+ a^{2}b^{2}(a + b)^{2}\(\frac{1}{b^{2}}\)

a^{2}b^{2} = b^{2}(a + b)^{2} + a^{2}(a + b)^{2}

a^{2}b^{2} = (a^{2} + b^{2})(a + b)^{2}

a^{2}b^{2} = a^{4} + 2a^{3}b + 2a^{2}b^{2} + 2ab^{3} + b^{4}

a^{2}b^{2} = a^{4} + 2ab(a^{2} + b^{2}) + 2a^{2}b^{2} + b^{4}

0 = a^{4} + 2ab(a^{2} + b^{2}) + a^{2}b^{2} + b^{4}

The terms a^{4}, a^{2}b^{2}, and b^{4} are positive because they are even-numbered powers of nonzero numbers. We established in the lesson that ab = (a + b)^{2} and, therefore, is also positive.

The product 2ab(a^{2} + b^{2}) must then also be positive.

The sum of four positive numbers will never equal zero, so there are no real solutions for a and b.

Question 6.

Let f(x) = log(x). Find values of a such that f(3a) = 3f(a).

Answer:

log(3a) = 31og(a)

log(3a) = log(a)^{3}

3a = a^{3}

3 = a^{2}

a = \(\sqrt{3}\)

This is true for the value of a when 3a = a^{3} that is in the domain, which is a = \(\sqrt{3}\).

Question 7.

Let f(x) = log(x). Find values of a such that f(ka) = kf(a).

Answer:

This is true for the values of a when ka = ak that are in the domain of the function.

Question 8.

Based on your results from the previous two problems, form a conjecture about whether f(x) = log(x) represents a linear transformation.

Answer:

The function is not an example of a linear transformation. The condition f(ka) = kf(a) does not hold for all values of a, for example, nonzero values of c and a = 1.

Question 9.

Let f(x) = ax^{2} + bx + c.

a. Describe the set of all values for a, b, and c that make f(x + y)= f(x) + f(y) valid for all real numbers x and y.

Answer:

This is challenging for students, but the goal is for them to realize that a = 0, c = 0, and any real number b. They may understand that a = 0, but c = 0 could be more challenging. The point is that it is unusual for functions to satisfy this condition for all real values of x and y. This is discussed in detail in Lesson 3.

f(x + y) = a(x + y)^{2} + b(x + y) + c = ax^{2} + 2axy + ay^{2} + bx + by + c

f(x) + f(y) = ax^{2} + bx + c + ay^{2} + by + c

f(x + y)= f(x) + f(y)

ax^{2} + 2axy + ay^{2} + bx + by + c = ax^{2} + bx + c + ay^{2} + by + c

2axy + c = c + c

2axy = c

Therefore, the set of values that satisfies this equation for all real numbers x and y is a = 0, any real number b, and c = 0.

b. What does your result indicate about the linearity of quadratic functions?

Answer:

Answers will vary but should address that quadratic functions are not linear transformations, since they only meet the condition f(x + y) = f(x) + f(y) when a = 0.

Trigonometry Table

Answer:

### Eureka Math Precalculus Module 1 Lesson 2 Exit Ticket Answer Key

Question 1.

Koshi says that he knows that sin(x + y) = sin(x) + sin(y) because he has substituted in multiple values for x and y, and they all work. He has tried x = 0Â° and y = 0Â°, but he says that usually works, so he also tried x = 45Â° and y = 180Â°, x = 90Â° and y = 270Â°, and several others. Is Koshi correct? Explain your answer.

Answer:

Koshi is not correct. He happened to pick values that worked, most giving at least one value of sin(x) = 0. If he had chosen other values such as x = 30Â° and y = 60Â°, sin(30Â° + 60Â°) = sin(90Â°) = 1, but sin(30Â°) + sin(60Â°) = \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\), so the statement that sin(30Â° + 60Â°) = sin(30Â°) + sin(60Â°) is false.

Question 2.

Is f(x) = sin(x) a linear transformation? Why or why not?

Answer:

No. sin(x + y) â‰ sin(x) + sin(y) and sin(ax) â‰ a sin(x)