Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 3.2
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Exercise 116 Page 158

What values of x can be used as input for the function? g(-5) and g(a+1) are the function values when x=-5 and x=a+1. The equations g(x)=32 and g(x)=0 are quadratic equations.

Domain: All real numbers
Range: g(x)≥ 0
Values:
g(-5)=8
g(a+1)=2a^2+16a+32
g(x)=32 for x=-7 and x=1
g(x)=0 for x=-3

Practice makes perfect

Here we have many questions in the same exercise. Let's answer them one at the time.

Domain

The domain is all x-values than can be used as input for the function. There is no x-value where the expression becomes undefined. Therefore, the domain is all real numbers. All real numbers

Range

The range is all function values that can be obtained. Consider the expression (x+3)^2. A squared number is always greater than or equal to 0. Therefore, 2(x+3)^2 is always greater than or equal to 0, no matter what x is. Therefore, the range can be described as follows.

g(x)≥ 0

Output Values

The expression g(-5) and g(a+1) is the function value when x=-5 and x=a+1. Let's substitute this value and expression for x and evaluate.

x 2(x+3)^2 g(x)
-5 2( -5+3)^2 8
To calculate g(a+1), we need to perform a few more calculation steps.
g(x)=2(x+3)^2
g( a+1)=2(( a+1)+3)^2
â–Ľ
Simplify right-hand side
g(a+1)=2(a+1+3)^2
g(a+1)=2(a+4)^2
g(a+1)=2(a^2+8a+16)
g(a+1)=2a^2+16a+32

Input Values

The solution to the equation g(x)=32 is the x-value(s) that make the function equal 32. Therefore, we have to set g(x) equal to 32 and solve for x.
g(x)=2(x+3)^2
32=2(x+3)^2
â–Ľ
Solve for x
16=(x+3)^2
(x+3)^2=16
x+3=±sqrt(16)
x+3=±4
x=-3±4
lcx=-3-4 & (I) x=-3+4 & (II)

(I), (II): Add and subtract terms

lx_1=-7 x_2=1
To solve g(x)=0, we have to set g(x) equal to 0 and solve for x.
g(x)=2(x+3)^2
0=2(x+3)^2
â–Ľ
Solve for x
0=(x+3)^2
(x+3)^2=0
x+3=0
x=-3
The solution to the equation is x=-3.