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Core Connections Integrated I, 2013

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Core Connections Integrated I, 2013 View details

2. Section 11.2

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Exercise 129 Page 627

a To calculate

f (- 1)

we have to substitute

x = - 1

into the function and simplify the right-hand side.

f (x) = ∣ x - 3 ∣ + 5

$x = - 1$

f (- 1) = ∣ - 1 - 3 ∣ + 5

Subtract term

f (- 1) = ∣ - 4 ∣ + 5

$∣ - 4 ∣ = 4$

f (- 1) = 4 + 5

Add terms

f (- 1) = 9

b Like in Part A, we have to substitute

x

with what is inside the parentheses and simplify the right-hand side.

f (x) = ∣ x - 3 ∣ + 5

$x = 3 - 8$

f (3 - 8) = ∣ ∣ ∣ 3 - 8 - 3 ∣ ∣ ∣ + 5

Calculate root

f (3 - 8) = ∣ - 2 - 3 ∣ + 5

Subtract term

f (3 - 8) = ∣ - 5 ∣ + 5

$∣ - 5 ∣ = 5$

f (3 - 8) = 5 + 5

Add terms

f (3 - 8) = 10

c In this case, we want to find the corresponding value of

x

when

f (x)

is

10 .

Therefore, we substitute

f (x) = 10

and solve for

x .

f (x) = ∣ x - 3 ∣ + 5

$f (x) = 10$

10 = ∣ x - 3 ∣ + 5

$LHS - 5 = RHS - 5$

5 = ∣ x - 3 ∣

Rearrange equation

∣ x - 3 ∣ = 5

When we remove the absolute value from the left-hand side we get two equations. One equation where the right-hand side is positive, and one where the right-hand side is negative.

∣ x - 3 ∣ = 5 ⇓ x - 3 = 5 x - 3 = - 5

Let's solve these one at the time.

x - 3 = 5

$LHS + 3 = RHS + 3$

x = 8

Let's also solve the last equation.

x - 3 = - 5

$LHS + 3 = RHS + 3$

x = - 2

Subchapter links

11.2.1 Solving Work and Mixture Problems p.603-604

11.2.2 Solving Equations and Systems Graphically p.607-608

11.2.3 Using a Best-Fit Line to Make a Prediction p.610-614

11.2.4 Treasure Hunt p.616-617

11.2.5 Using Coordinate Geometry and Constructions to Explore Shapes p.620-621

11.2.6 Modeling with Exponential Functions and Linear Inequalities p.625-628