Eureka Math Precalculus Module 3 Lesson 11 Answer Key

Engage NY Eureka Math Precalculus Module 3 Lesson 11 Answer Key

Eureka Math Precalculus Module 3 Lesson 11 Example Answer Key

Example 1.
Simplify the expression \(\) to lowest terms, and identify the value(s) of x that must be excluded to avoid division by zero.
Answer:
Since x2-5x + 6 = (x-2)(x-3), our original expression can be written as \(\frac{x^{2}-5 x + 6}{x-3}\) = \(\frac{(x-2)(x-3)}{x-3}\). To simplify this expression to lowest terms, we need to divide the numerator and denominator by any common factors. The only common factor in this example is x-3. However, we can only divide by x-3 if x-3 ≠ 0, which means that we have to exclude 3 as a possible value of x.
Thus, if x ≠ 3, we have
\(\frac{x^{2}-5 x + 6}{x-3}\) = \(\frac{(x-2)(x-3)}{x-3}\) =
= \(\frac{(x-2)(x-3)}{x-3}\)⋅\(\frac{\frac{1}{x-3}}{\frac{1}{x-3}}\)
= x-2.
So, as long as x ≠ 3, the expressions \(\frac{x^{2}-5 x + 6}{x-3}\) and x-2 are equivalent.

Example 2.
Let f(x) = \(\frac{2 x^{4} + 6 x^{3} + 6 x^{2} + 2 x}{3 x^{2} + 3 x}\). Simplify the rational expression \(\frac{2 x^{4} + 6 x^{3} + 6 x^{2} + 2 x}{3 x^{2} + 3 x}\) to lowest terms, and use the simplified form to express the rule of f. Be sure to indicate any restrictions on the domain.
Answer:
\(\frac{2 x^{4} + 6 x^{3} + 6 x^{2} + 2 x}{3 x^{2} + 3 x}\) = \(\frac{2 x\left(x^{3} + 3 x^{2} + 3 x + 1\right)}{3 x(x + 1)}\) = \(\frac{2 x(x + 1)^{3}}{3 x(x + 1)}\) = \(\frac{2(x + 1)^{2}}{3}\) if x ≠ -1 and x ≠ 0.
Then, f(x) = \(\frac{2(x + 1)^{2}}{3}\) for x ≠ 0 and x ≠ -1.

Eureka Math Precalculus Module 3 Lesson 11 Exercise Answer Key

Opening Exercise
Factor each expression completely:
a. 9x4-16x2
Answer:
x2 (3x + 4)(3x-4)

b. 2x3 + 5x2-8x-20
Answer:
(x2-4)(2x + 5) = (x + 2)(x-2)(2x + 5)

c. x3 + 3x2 + 3x + 1
Answer:
(x + 1)3

d. 8x3-1
Answer:
(2x-1)(4x2 + 2x + 1)

Exercise 1: Simplifying Rational Expressions to Lowest Terms
Exercise 1.
Simplify each rational expression to lowest terms, specifying the values of x that must be excluded to avoid division by zero.
a. \(\frac{x^{2}-6 x + 5}{x^{2}-3 x-10}\)
Answer:
The denominator factors into x2-3x-10 = (x-5)(x + 2), so to avoid division by zero, we must have
x ≠ 5 and x ≠ -2. Thus, \(\frac{x^{2}-6 x + 5}{x^{2}-3 x-10}\) = \(\frac{(x-5)(x-1)}{(x-5)(x + 2)}\) = \(\frac{x-1}{x + 2}\), where x ≠ -2 and x ≠ 5.

b. \(\frac{x^{3} + 3 x^{2} + 3 x + 1}{x^{3} + 2 x^{2} + x}\)
The denominator factors into x3 + 2x2 + x = x(x + 1)2, so to avoid division by zero, we must have x ≠ 0 and x ≠ -1. Thus, \(\frac{x^{3} + 3 x^{2} + 3 x + 1}{x^{3} + 2 x^{2} + x}\) = \(\frac{(x + 1)^{3}}{x(x + 1)^{2}}\) = \(\frac{x + 1}{x}\), where x ≠ 0 and x ≠ 1.

c. \(\frac{x^{2}-16}{x^{2} + 2 x-8}\)
The denominator factors into x2 + 2x-8 = (x-2)(x + 4), so to avoid division by zero, we must have x ≠ 2 and x ≠ -4. Thus, \(\frac{x^{2}-16}{x^{2} + 2 x-8}\) = \(\frac{(x-4)(x + 4)}{(x-2)(x + 4)}\) = \(\frac{x-4}{x-2}\), where x ≠ 2 and x ≠ -4.

d. \(\frac{x^{2}-3 x-10}{x^{3} + 6 x^{2} + 12 x + 8}\)
Answer:
The denominator factors into x3 + 6x2 + 12x + 8 = (x + 2)3, so to avoid division by zero, we must have x ≠ -2. Thus, \(\frac{x^{2}-3 x-10}{x^{3} + 6 x^{2} + 12 x + 8}\) = \(\frac{(x-5)(x + 2)}{(x + 2)^{3}}\) = \(\frac{x-5}{(x + 2)^{2}}\), where x ≠ -2.

e. \(\frac{x^{3} + 1}{x^{2} + 1}\)
Answer:
While x3 + 1 = (x + 1)(x2-x + 1), the polynomial expression in the denominator does not factor. Thus, this expression is already simplified to lowest terms. Since x2 + 1≥1 for all values of x, the denominator is never zero. Thus, there are no values of x that need to be excluded.

Exercise 2.
Determine the domain of each rational function, and express the rule for each function in an equivalent form in lowest terms.
a. f(x) = \(\frac{(x + 2)^{2}(x-3)(x + 1)}{(x + 2)(x + 1)}\)
Answer:
The domain is all real numbers x so that x ≠ -1 and x ≠ -2.
f(x) = \(\frac{(x + 2)^{2}(x-3)(x + 1)}{(x + 2)(x + 1)}\) = (x + 2)(x-3) for x ≠ -1 and x ≠ -2

b. f(x) = \(\frac{x^{2}-6 x + 9}{x-3}\)
Answer:
The domain is all real numbers x so that x ≠ 3.
f(x) = \(\frac{(x-3)^{2}}{x-3}\) = x-3 for x ≠ 3

c. f(x) = \(\frac{3 x^{3}-75 x}{x^{3} + 15 x^{2} + 75 x + 125}\)
Answer:
The domain is all real numbers x so that x ≠ -5.
f(x) = \(\frac{3 x\left(x^{2}-25\right)}{(x + 5)^{3}}\) = \(\frac{3 x(x + 5)(x-5)}{(x + 5)^{3}}\) = \(\frac{3 x(x-5)}{(x + 5)^{2}}\) for x ≠ -5

Eureka Math Precalculus Module 3 Lesson 11 Problem Set Answer Key

Question 1.
For each pair of functions f and g, find the domain of f and the domain of g. Indicate whether f and g are the same function.
a. f(x) = \(\frac{x^{2}}{x}\), g(x) = x
Answer:
The domain of f is all real numbers x with x ≠ 0. The domain of g is all real numbers x.
No, functions f and g are not the same function.

b. f(x) = \(\frac{x}{x}\), g(x) = 1
Answer:
The domain of f is all real numbers x with x ≠ 0. The domain of g is all real numbers x.
No, functions f and g are not the same function.

c. f(x) = \(\frac{2 x^{2} + 6 x + 8}{2}\), g(x) = x2 + 6x + 8
Answer:
The domain of f is all real numbers x. The domain of g is all real numbers x.
Yes, functions f and g are the same function.

d. f(x) = \(\frac{x^{2} + 3 x + 2}{x + 2}\), g(x) = x + 1
Answer:
The domain of f is all real numbers x with x ≠ -2. The domain of g is all real numbers x.
No, functions f and g are not the same function.

e. f(x) = \(\frac{x + 2}{x^{2} + 3 x + 2}\), g(x) = \(\frac{1}{x + 1}\)
Answer:
The domain of f is all real numbers x with x ≠ -2 and x ≠ -1. The domain of g is all real numbers x with x ≠ -1.
No, functions f and g are not the same function.

f. f(x) = \(\frac{x^{4}-1}{x^{2}-1}\), g(x) = x2 + 1
Answer:
The domain of f is all real numbers x with x ≠ 1 and x ≠ -1. The domain of g is all real numbers x.
No, functions f and g are not the same function.

g. f(x) = \(\frac{x^{4}-1}{x^{2} + 1}\), g(x) = x2-1
Answer:
Because x2 + 1 is never zero, the domain of f is all real numbers x. The domain of g is all real numbers x.
Yes, functions f and g are the same function.

h. f(x) = \(\frac{x^{4}-x}{x^{2} + x}\), g(x) = \(\frac{x^{3}-1}{x + 1}\)
Answer:
The domain of f is all real numbers x with x ≠ 0 and x ≠ -1. The domain of g is all real numbers x with x ≠ -1.
No, functions f and g are not the same function.

i. f(x) = \(\frac{x^{4} + x^{3} + x^{2}}{x^{2} + x + 1}\), g(x) = x2
Answer:
Because x2 + x + 1 doesn’t factor, the denominator of f is never zero, and the domain of f is all real numbers x. The domain of g is also all real numbers x.
Yes, functions f and g are the same function.

Question 2.
Determine the domain of each rational function, and express the rule for each function in an equivalent form in lowest terms.
a. f(x) = \(\frac{x^{4}}{x^{2}}\)
Answer:
The domain of f is all real numbers x with x ≠ 0.
f(x) = \(\frac{x^{4}}{x^{2}}\) = x2, where x ≠ 0

b. f(x) = \(\frac{3 x + 3}{15 x-6}\)
Answer:
Because 15x-6 = 3(5x-2), the domain of f is all real numbers x with x ≠ \(\frac{2}{5}\).
f(x) = \(\frac{3(x + 1)}{3(5 x-2)}\) = \(\frac{x + 1}{5 x-2}\), where x ≠ \(\frac{2}{5}\)

c. f(x) = \(\frac{x^{2}-x-2}{x^{2} + x}\)
Answer:
Because x2 + x = x(x + 1), the domain of f is all real numbers x with x ≠ 0 and x ≠ -1.
f(x) = \(\frac{x^{2}-x-2}{x^{2} + x}\) = \(\frac{(x-2)(x + 1)}{x(x + 1)}\) = \(\frac{x-2}{x}\), where x ≠ 0 and x ≠ -1

d. f(x) = \(\frac{8 x^{2} + 2 x-15}{4 x^{2}-4 x-15}\)
Answer:
Because 4x2-4x-15 = (2x + 3)(2x-5), the domain of f is all real numbers x with x ≠ –\(\frac{3}{2}\) and x ≠ \(\frac{5}{2}\).
f(x) = \(\frac{8 x^{2} + 2 x-15}{4 x^{2}-4 x-15}\) = \(\frac{(2 x + 3)(4 x-5)}{(2 x + 3)(2 x-5)}\) = \(\frac{(4 x-5)}{(2 x-5)}\), where x ≠ \(\frac{5}{2}\) and x ≠ –\(\frac{3}{2}\)

e. f(x) = \(\frac{2 x^{3}-3 x^{2}-2 x + 3}{x^{3}-x}\)
Answer:
Because x3-x = x(x-1)(x + 1), the domain of f is all real numbers x with x ≠ 0, x ≠ 1 and x ≠ -1.
f(x) = \(\frac{2 x^{3}-3 x^{2}-2 x + 3}{x^{3}-x}\) = \(\frac{(2 x-3)\left(x^{2}-1\right)}{x\left(x^{2}-1\right)}\) = \(\frac{2 x-3}{x}\), where x ≠ 0 and x ≠ ±1

f. f(x) = \(\frac{3 x^{3} + x^{2} + 3 x + 1}{x^{3} + x}\)
Answer:
Because x3 + x = x(x2 + 1), the domain of f is all real numbers x with x ≠ 0.
f(x) = \(\frac{3 x^{3} + x^{2} + 3 x + 1}{x^{3} + x}\) = \(\frac{(3 x + 1)\left(x^{2} + 1\right)}{x\left(x^{2} + 1\right)}\) = \(\frac{3 x + 1}{x}\), where x ≠ 0

Question 3.
For each pair of functions below, calculate f(x) + g(x), f(x)-g(x), f(x)⋅g(x), and \(\frac{f(x)}{g(x)}\). Indicate restrictions on the domain of the resulting functions.
a. f(x) = \(\frac{2}{x}\), g(x) = \(\frac{x}{x + 2}\)
Answer:
Eureka Math Precalculus Module 3 Lesson 11 Problem Set Answer Key 1

b. f(x) = \(\frac{3}{x + 1}\), g(x) = \(\frac{x}{x^{3} + 1}\)
Answer:
Eureka Math Precalculus Module 3 Lesson 11 Problem Set Answer Key 2

Eureka Math Precalculus Module 3 Lesson 11 Exit Ticket Answer Key

Question 1.
Identify whether the functions shown are rational:
a. f(x) = \(\frac{x}{x^{2} + 1}\)
Answer:
Yes. Both P(x) = x and Q(x) = x2 + 1 are polynomial functions.

b. f(x) = \(\frac{\sqrt{\boldsymbol{x}}}{x^{2} + 1}\)
Answer:
No. The function P(x) = \(\sqrt{x}\) is not a polynomial function.

c. f(x) = \(\frac{x}{x^{0.4} + 1}\)
Answer:
No. The function Q(x) = x(0.4) + 1 is not a polynomial function.

d. f(x) = (\(\frac{x}{x^{2} + 1}\))2
Answer:
Yes. When multiplied out, f(x) = \(\frac{x^{2}}{x^{4} + 2 x^{2} + 1}\), so f is the quotient of two polynomial functions.

e. f(x) = \(\frac{\sqrt{2} x}{e x^{2} + \sqrt{\pi}}\)
Answer:
Yes. While the coefficients are not integers, P(x) = \(\sqrt{2}\) x and Q(x) = ex2 + \(\sqrt{\pi}\) are both polynomial functions since all the powers of x are whole numbers.

Question 2.
Anmol says f(x) = \(\frac{x + 1}{x^{2}-1}\) and g(x) = \(\frac{1}{x-1}\)represent the same function. Is she correct? Justify your answer.
Answer:
She is not correct.
The function f(x) = \(\frac{x + 1}{x^{2}-1}\) is not defined for x = 1 and x = -1. However, the function g(x) = \(\frac{1}{x-1}\) is not defined for x = 1. These two functions do not have the same domain, so they are not the same function.

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