# Eureka Math Precalculus Module 1 Lesson 2 Answer Key

## 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.
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.
log(2a) = 21og(a)
log(2a) = log(a2)
2a = a2
a2 – 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) = 10x. 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.
For a = b = log(2)
f(a + b) = 1o(log(2)+log(2)) = 102log(2) = 10log(22) = 22 = 4
f(a) + f(b) = 10log2 + 10log2 = 2 + 2 = 4
Therefore, f(a + b) = f(a) + f(b).
For a = b = 2
f(a + b) = 102+2 = 104 = 10000
f(a) + f(b) = 102 + 102 = 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.
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 = a2 + ab + ab + b2
ab = a2 + 2ab + b2
ab = (a + b)2
This means that ab must be a positive number. Simplifying further, we get 0 = a2 + ab + b2.
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.
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)
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)
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.

Question 3.
tan(x + y) = tan(x) + tan(y)
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)
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.
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 a2b2(a + b)2, we get
a2b2(a + b)2$$\frac{1}{(a+b)^{2}}$$= a2b2(a + b)2$$\frac{1}{a^{2}}$$+ a2b2(a + b)2$$\frac{1}{b^{2}}$$
a2b2 = b2(a + b)2 + a2(a + b)2
a2b2 = (a2 + b2)(a + b)2
a2b2 = a4 + 2a3b + 2a2b2 + 2ab3 + b4
a2b2 = a4 + 2ab(a2 + b2) + 2a2b2 + b4
0 = a4 + 2ab(a2 + b2) + a2b2 + b4
The terms a4, a2b2, and b4 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(a2 + b2) 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).
log(3a) = 31og(a)
log(3a) = log(a)3
3a = a3
3 = a2
a = $$\sqrt{3}$$
This is true for the value of a when 3a = a3 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).
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.
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) = ax2 + 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.
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 = ax2 + 2axy + ay2 + bx + by + c
f(x) + f(y) = ax2 + bx + c + ay2 + by + c
f(x + y)= f(x) + f(y)
ax2 + 2axy + ay2 + bx + by + c = ax2 + bx + c + ay2 + 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.

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

### 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.
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.