Determine algebraically whether the function f(x)=2x^42x^2 2x+1 is odd, even or neither.
Even: Algebraically, we substitute x as -x and if it goes back to original function, then it's even. Thus, f(-x) = f(x).
Visually, if this function is graphed and you see that it is symmetric to y-axis, then it's even.
Odd: If f(-x) = -f(x).
I will assume the function you wrote is
f(x) = 2x^4 + 2x^2 + 2x + 1
Let x be -x.
f(x) = 2(-x)^4 + 2(-x)^2 + 2(-x) + 1
f(x) = 2x^4 + 2x^2 - 2x + 1
The resulting function is neither odd or even, since it didn't result in f(x) or -f(x).
To determine whether a function is odd, even, or neither, we need to examine the properties of the function in relation to its input (x) and output (f(x)) values.
1. For a function to be odd, it must satisfy the property: f(-x) = -f(x). This means that substituting -x for x in the function should yield the negative of the original function.
2. For a function to be even, it must satisfy the property: f(-x) = f(x). This means that substituting -x for x in the function should yield the same value as the original function.
Now let's apply these properties to the function f(x) = 2x^4 + 2x^2 + 2x + 1:
1. Checking for oddness:
f(-x) = 2(-x)^4 + 2(-x)^2 + 2(-x) + 1
= 2x^4 + 2x^2 - 2x + 1
To check if f(-x) = -f(x), we compare this with the original function:
-f(x) = -(2x^4 + 2x^2 + 2x + 1)
= -2x^4 - 2x^2 - 2x - 1
We can see that f(-x) does not equal -f(x). Therefore, the function is not odd.
2. Checking for evenness:
f(-x) = 2(-x)^4 + 2(-x)^2 + 2(-x) + 1
= 2x^4 + 2x^2 - 2x + 1
To check if f(-x) = f(x), we compare this with the original function:
f(x) = 2x^4 + 2x^2 + 2x + 1
We can see that f(-x) equals f(x). Therefore, the function is even.
In conclusion, the function f(x) = 2x^4 + 2x^2 + 2x + 1 is an even function.