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