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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
Here we have many questions in the same exercise. Let's answer them one at the time.
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
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.
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.
x= a+1
Remove parentheses
Add terms
(a+b)^2=a^2+2ab+b^2
Distribute 2
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.
To solve g(x)=0, we have to set g(x) equal to 0 and solve for x.
g(x)= 0
.LHS /2.=.RHS /2.
Rearrange equation
sqrt(LHS)=sqrt(RHS)
LHS-3=RHS-3
The solution to the equation is x=-3.