To classify the symmetry of the function [tex]\( f(x) = x^4 + x^2 - 1 \)[/tex], we need to determine if it is even, odd, or neither. We'll do this by considering the definitions of even and odd functions:
1. A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(x) = f(-x) \)[/tex] for all [tex]\( x \)[/tex] in the domain of [tex]\( f \)[/tex].
2. A function [tex]\( f(x) \)[/tex] is odd if [tex]\( f(x) = -f(-x) \)[/tex] for all [tex]\( x \)[/tex] in the domain of [tex]\( f \)[/tex].
Let's proceed with checking both possible symmetries:
### 1. Checking for Even Symmetry
A function is even if [tex]\( f(x) = f(-x) \)[/tex]. Let's substitute [tex]\( -x \)[/tex] into the function and see if it equals [tex]\( f(x) \)[/tex]:
[tex]\[
f(-x) = (-x)^4 + (-x)^2 - 1
\][/tex]
Simplify the terms:
[tex]\[
= x^4 + x^2 - 1
\][/tex]
We see that:
[tex]\[
f(-x) = f(x)
\][/tex]
This shows that [tex]\( f(x) \)[/tex] satisfies the condition for even symmetry.
### 2. Checking for Odd Symmetry
A function is odd if [tex]\( f(x) = -f(-x) \)[/tex]. Let's substitute [tex]\( -x \)[/tex] into the function and check if the equation holds:
[tex]\[
f(-x) = (-x)^4 + (-x)^2 - 1
\][/tex]
Simplify the terms again:
[tex]\[
= x^4 + x^2 - 1
\][/tex]
Now, we need to see if:
[tex]\[
f(x) = -f(-x)
\][/tex]
This would mean:
[tex]\[
x^4 + x^2 - 1 = -(x^4 + x^2 - 1)
\][/tex]
Which simplifies to:
[tex]\[
x^4 + x^2 - 1 = -x^4 - x^2 + 1
\][/tex]
Clearly, this equality does not hold true.
Thus, [tex]\( f(x) \)[/tex] is not an odd function.
### Conclusion
Since [tex]\( f(x) = x^4 + x^2 - 1 \)[/tex] satisfies the condition for even symmetry and not the odd symmetry, we conclude that the function [tex]\( f(x) \)[/tex] is even.
The correct classification of the symmetry for the function [tex]\( f(x) = x^4 + x^2 - 1 \)[/tex] is:
even.
