f(x)=x^3 + sin x
So far.. I've heard it's neither.. But how? Thanks
to be odd, f(-x)=-f(x)
does f(-100)=-f(100)
f(-100)=-(100)^3 +sin(-100)
=-(100)^3-sin(100)
f(100)=100^3+sin(100)
so does f(-x)=-f(x) ?
f(x)=2/π+4/π(cos2x/3-cos4x/15+....)
To determine whether the function f(x) = x^3 + sin(x) is even, odd, or neither, we need to evaluate f(-x) and compare it with f(x).
First, let's evaluate f(-x):
f(-x) = (-x)^3 + sin(-x)
= -x^3 + sin(-x)
Now, let's compare f(x) with f(-x):
f(x) = x^3 + sin(x)
f(-x) = -x^3 + sin(-x)
If f(x) = f(-x), then the function is even.
If f(x) = -f(-x), then the function is odd.
If neither condition is met, then the function is neither even nor odd.
Let's check these conditions for f(x) = x^3 + sin(x):
Condition 1: f(x) = f(-x)
x^3 + sin(x) = -x^3 + sin(-x)
Since sin(-x) = -sin(x), we have:
x^3 + sin(x) = -x^3 - sin(x)
This equation is only true if x = 0. However, for an even function, f(-x) should be equal to f(x) for all values of x, not just a particular value. Therefore, condition 1 is not met.
Condition 2: -f(-x) = f(x)
-( -x^3 + sin(-x) ) = x^3 + sin(x)
x^3 - sin(-x) = x^3 + sin(x)
Since sin(-x) = -sin(x), we have:
x^3 + sin(x) = x^3 + sin(x)
This equation is true for all values of x. Therefore, condition 2 is met.
Based on these conditions, we can conclude that f(x) = x^3 + sin(x) is an odd function.