According to the visible range of the graph, is f(x) even, odd, or neither?
f(x) is graphed below.
f(x) = f(-x) for all values of x.f(x) = -f(-x) for all values of x.
\qquad f(\blue{PT}) \approx \red{roundTo(1, FUNC(PT))}\qquad f(\blue{-PT}) \approx \red{roundTo(1, FUNC(-PT))}f(\blue{PT}) \neq f(\blue{-PT}), so f(x) is not even.
f(\blue{PT}) \neq -f(\blue{-PT}), so f(x) is not odd.
Therefore f(x) is neither.
\qquad f(\blue{x}) \approx \red{roundTo(1, FUNC(x))}\qquad f(\blue{-x}) \approx \red{roundTo(1, FUNC(-x))}f(x) is odd because f(x) = -f(-x). (For all x values, not just the ones we checked!)
f(x) is even because f(x) = f(-x). (For all x values, not just the ones we checked!)
Is f(x) even, odd, or neither?
f(x) = FUNC
f(-x) = f(x) for all values of x.f(-x) = -f(x) for all values of x.What is f(\blue{-x})?
f(\blue{-x}) = NEG_FUNC1
f(\blue{-x}) = NEG_FUNC2
f(\blue{-x}) = -(FUNC)
f(\blue{-x}) = -f(x)
Therefore, f(x) is odd.
f(\blue{-x}) = f(x)
Therefore, f(x) is even.
f(-x) \neq f(x), since the signs of the terms with odd powers are different.
f(-x) \neq -f(x) since the signs of the terms with even powers are the same.
Therefore, f(x) is neither odd nor even.
SOL_TEXT