f(x) = |x + 1| + |x − 1|
To prove whether the function is odd or even, you have to test it for symmetry.
First, test whether it's even by seeing if the function is symmetrical about the y-axis.
Remember that in order for that kind of symmetry to occur, the following must be true for all values of x:
f(-x) = f(x)
So, trying it:
f(-x) = |(-x) + 1| + |(-x) − 1|
f(-x) = |(-1)(x − 1)| + |(-1)(x + 1)| ... Factor out the (-1).
f(-x) = |x − 1| + |x + 1| ... We get the function, itself, just rearranged.
f(-x) = f(x) ... It satisfies.
Therefore, f(x) is symmetrical about the y-axis and is even.
Next, testing to see if it is odd requires that the function is symmetrical about the origin, accomplished by the identity:
f(-x) = -f(x)
This is impossible for the given function, since testing negative x already gave us the function, itself, which is not equal to its negative value.
Therefor, the function f(x) is even, and only even.
~~~
x + 1 < 0 => |x+1| + |x-1| = -x - 1 - x + 1 = -2x which is even only when x is a whole number
x between -1 and 0 => |x+1| + |x-1| = x + 1 - x + 1 = 2 for all x
0 <= x < 1 => |x+1| + |x-1| = x + 1 - x + 1 = 2 for all x
x >= 1 => |x+1| + |x-1| = 2x which is even only when x a whole number
summary:
|x+1| + |x-1| is 2 when x between -1 and 1 whether x is whole or not
everywhere else it is 2x which is even only for whole numbers
if x is a whole number, |x+1| + |x-1| is a whole even number
To find out if a function is odd or even all you do is plug in -x to where all the x's are
F(-x)=|-x+1|+|-x-1|
If you solve and you get the same function it's even, if you get out -(the same function) its odd and if u get a mix of signs it's neither
If x is odd, you get even+even, which is even. If you plug in an even number for x, then you get odd+odd, which is also even. This will always be even.
Even
f(-x)= |(-x)+1| +|(-x)-1|
=|-x+1|+|-x-1|
we have |-x-1|=|x+1| (|-x)=|x|)
and |-x+1|=|x-1|
so f(-x)=|x+1|+|x-1|=f(x)
so f is even
it will always be even since there are absoulte value
it's surely even
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Verified answer
f(x) = |x + 1| + |x − 1|
To prove whether the function is odd or even, you have to test it for symmetry.
First, test whether it's even by seeing if the function is symmetrical about the y-axis.
Remember that in order for that kind of symmetry to occur, the following must be true for all values of x:
f(-x) = f(x)
So, trying it:
f(-x) = |(-x) + 1| + |(-x) − 1|
f(-x) = |(-1)(x − 1)| + |(-1)(x + 1)| ... Factor out the (-1).
f(-x) = |x − 1| + |x + 1| ... We get the function, itself, just rearranged.
f(-x) = f(x) ... It satisfies.
Therefore, f(x) is symmetrical about the y-axis and is even.
Next, testing to see if it is odd requires that the function is symmetrical about the origin, accomplished by the identity:
f(-x) = -f(x)
This is impossible for the given function, since testing negative x already gave us the function, itself, which is not equal to its negative value.
Therefor, the function f(x) is even, and only even.
~~~
x + 1 < 0 => |x+1| + |x-1| = -x - 1 - x + 1 = -2x which is even only when x is a whole number
x between -1 and 0 => |x+1| + |x-1| = x + 1 - x + 1 = 2 for all x
0 <= x < 1 => |x+1| + |x-1| = x + 1 - x + 1 = 2 for all x
x >= 1 => |x+1| + |x-1| = 2x which is even only when x a whole number
summary:
|x+1| + |x-1| is 2 when x between -1 and 1 whether x is whole or not
everywhere else it is 2x which is even only for whole numbers
if x is a whole number, |x+1| + |x-1| is a whole even number
To find out if a function is odd or even all you do is plug in -x to where all the x's are
F(-x)=|-x+1|+|-x-1|
If you solve and you get the same function it's even, if you get out -(the same function) its odd and if u get a mix of signs it's neither
If x is odd, you get even+even, which is even. If you plug in an even number for x, then you get odd+odd, which is also even. This will always be even.
Even
f(-x)= |(-x)+1| +|(-x)-1|
=|-x+1|+|-x-1|
we have |-x-1|=|x+1| (|-x)=|x|)
and |-x+1|=|x-1|
so f(-x)=|x+1|+|x-1|=f(x)
so f is even
it will always be even since there are absoulte value
it's surely even