I’m confused with question 14 from the algebra foundation quiz #1.
I understand why choice B and D are even functions, but I don’t understand why choice A is not even.
This is choice A: f(x) = | x + 2|
Since f(-x) = f(x) , won’t giving a negative input to choice A always generate a positive output regardless? It’s like f(x) = |x| , but wouldn’t the constant not matter here?
If you take x=1, f(x) will be equal to |1+2| = 3
So f(-x) will be equal to |-1+2| = 1
As f(x) not equal to f(-x) the function is not even.
I see, so there must be some kind of “equivalence”? Like, f(-1)=1 or f(-10)=10 are okay but f(-1)=2 or f(-10)=11 are not.
No - it’s just f(x) = f(-x) for all real x.
Okay, I think I get it now.
A function is even when it satisfies this f(-x) = f(x) .
So let’s test that with f(x)=|x+2|
If x=1, f(1)=|1+2|=3
If x=-1, f(-1)=|-1+2|=1
f(1) \neq f(-1), because 3 \neq 1, hence it’s not even.
Let’s test this with choice B and D too.
If x=1, f(1)=|1+7|=8
If x=-1, f(-1)=|1+7|=8
f(1) = f(-1), because 8=8, hence it’s even.
If x=1, f(1)=|1|=1
If x=-1, f(-1)=|-1|=1
f(1) = f(-1), because 1=1, hence it’s even.
