# Eureka Math Precalculus Module 1 Lesson 1 Answer Key

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

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

Look at these common mistakes that students make, and answer the questions that follow.

Exercise 1.
If f(x) = $$\sqrt{x}$$, does f(a + b) = f(a) + f(b), when a and b are not negative?

a. Can we find a counterexample to refute the claim that f(a + b) = f(a) + f(b) for all nonnegative values of a and b?
Answers will vary, but students could choose a = 1 and b = 1. In this case, $$\sqrt{1+1}$$= $$\sqrt{1}$$ + $$\sqrt{1}$$, or $$\sqrt{2}$$ = 2, which we know is not true. There are many other choices.

b. Find some nonnegative values for a and b for which the statement, by coincidence, happens to be true.
Answers will vary but could include a = 0, b = 0 or a = 4, b = 0 or a = 0, b = 16.

c. Find all values of a and b for which the statement is true. Explain your work and the results.
$$\sqrt{a+b}$$ = $$\sqrt{a}$$ + $$\sqrt{b}$$
($$\sqrt{a+b}$$)2 = [$$\sqrt{a}$$ + $$\sqrt{b}$$)2
a + b = a + 2$$\sqrt{\boldsymbol{a b}}$$ + b
2$$\sqrt{\boldsymbol{a b}}$$ = 0
ab = 0,
which leads to a = 0 if students are solving for a and b = 0 if students are solving for b.
Anytime a = 0 and/or b = 0, then $$\sqrt{a+b}$$ = $$\sqrt{a}$$ + $$\sqrt{b}$$, and the equation is true.

d. Why was it necessary for us to consider only nonnegative values of a and b?
If either variable is negative, then we would be taking the square root of a negative number, which is not a real number, and we are only addressing real-numbered inputs and outputs here.

e. Does f(x) = $$\sqrt{\boldsymbol{x}}$$ display ideal linear properties? Explain.
No, because $$\sqrt{a+b}$$ ≠ $$\sqrt{a}$$ + $$\sqrt{b}$$ for all real values of the variables.

Question 2.
If f(x) = x3, does f(a + b) = f(a) + (b)?

a. Substitute in some values of a and b to show this statement is not true in general.
Answers will vary, but students could choose a = 1 and b = 1. In this case, (1 + 1)3 = 13 + 13 or 8 = 2, which we know is not true. There are many other choices.

b. Find some values for a and b for which the statement, by coincidence, happens to work.
Answers will vary but could include a = 0, b = 0 or a = 2, b = 0 or a = 0, b = 3.

c. Find all values of a and b for which the statement is true. Explain your work and the results.
(a + b)3 = a3 + b3
a3 + 3a2b + 3ab2 + b3 = a3 + b3
3a2b + 3ab2 = 0
3ab(a + ft) = 0,
which leads to a = 0, b = 0, and a = —b.
Anytime a = 0 and/or b = 0 or a = —b, then (a + b)3 = a3 + b3, and the equation is true.

d. Is this true for all positive and negative values of a and ft? Explain and prove by choosing positive and negative values for the variables.
Yes, since = —b, if a is positive, the equation would be true if b was negative. Likewise, if a is negative, the equation would be true if b was positive. Answers will vary. If a = 2 and b = —2,
(2 + (—2))3 = (2)3 + (—2)3, meaning 03 = 8 + (—8) or 0 = 0. If a = —2 and b = 2,
((—2) + 2)3 = (—2)3 + (2)3, meaning 03 = (—8) + 8, or 0 = 0. Therefore, this statement is true for all positive and negative values of a and b.

e. Does f(x) = x3 display ideal linear properties? Explain.
No, because (a + b)3 ≠ a3 + b3 for all real values of the variables.

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

Assign students some or all of the functions to investigate. All students should attempt Problem 4 to set up the next lesson. It is hoped that students may give some examples that are studied in Lesson 2.

Study the statements given in Problems 1-3. Prove that each statement is false, and then find all values of a and b for which the statement is true. Explain your work and the results.

Question 1.
If f(x) = x2, does f(a + b) = f(a) + f(b)?
Answers that prove the statement false will vary but could include a = 2 and b = —2.
This statement is true when a = 0 and/or b = 0.

Question 2.
If f(x) = $$x^{\frac{1}{3}}$$, does f(a + b) = f(a) + f(b)?
Answers that prove the statement false will vary but could include a = 1 and b = 1.
This statement is true when a = 0 and/or b = 0 and when a = —b.

Question 3.
If f(x) = $$\sqrt{4 x}$$, does f(a + b) = f(a) + f(b)?
Answers that prove the statement false will vary but could include a = —1 and b = 1.
This statement is true when a = 0 and/or b = 0.

Question 4.
Think back to some mistakes that you have made in the past simplifying or expanding functions. Write the statement that you assumed was correct that was not, and find numbers that prove your assumption was false.
Answers will vary but could include sin(x + y) = sin(x) + sin(y), which is false when x and y equal 45°, log(2a) = 2log(a), which is false for a = 1.
10a+b = 10a + 10b, which is false for a,b = 1, $$\frac{1}{a+b}$$ = $$\frac{1}{a}$$ + $$\frac{1}{b}$$, which is false for a, b = 1.

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

Question 1.
Xavier says that (a + b)2 ≠ a2 + b2 but that (a + b)3 = a3 + b3. He says that he can prove it by using the values a = 2 and b = -2. Shaundra says that both (a + b)2 = a2 + b2 and (a + b)3 = a3 + b3 are true and that she can prove it by using the values of a = 7 and b = 0 and also a = 0 and b = 3. Who is correct? Explain.
Neither is correct. Both have chosen values that just happen to work in one or both of the equations. In the first equation, anytime a = 0 and/or b = 0, the statement is true. In the second equation, anytime a = 0 and/or b = 0 and also when a = —b, the statement is true. If they tried other values such as a = 1 and b = 1, neither statement would be true.

Question 2.
Does f(x) = 3x + 1 display ideal linear properties? Explain.
No. f(ax) = 3ax + 1, but af(x) = 3ax + a These are not equivalent.
Also, f(x + y) = 3(x + y) + 1 = 3x + 3y + 1, but f(x) + f(y) = 3x + 1 + 3y + 1 = 3x + 3y + 2.
They are not equivalent, so the function does not display ideal linear properties.

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