How do you determine if #f(x)= x^3 + 1# is an even or odd function?

Question

Sadie Ackerman · Accepted Answer

neither

Examine the following to see if a function is even or odd. • f(x) is even if f(x) = f( -x) There is symmetry in functions about the y-axis. • f(x) is odd if f(-x) = - f(x). There is symmetry around the origin of odd functions. Check for even #f(-x)=(-x)^3+1=-x^3+1 ≠f(x)# f(x) is not even since f(x) ≠ f(-x). Check for odd #-f(x)=-(x^3+1)=-x^3-1≠f(-x)# f(x) is not odd since f(-x) ≠ - f(x). #rArrx^3+1" is neither even nor odd"# graph{x^3+1 [-10, 10, -5, 5]}

Caleb Alexander · Answer

undefined To determine if a function is even or odd, we need to examine its symmetry.

1. Even functions have symmetry about the y-axis, meaning that f(x) = f(-x) for all x in the domain. 
2. Odd functions have rotational symmetry of 180 degrees about the origin, meaning that f(x) = -f(-x) for all x in the domain.

For the function f(x) = x^3 + 1:
1. Check for even symmetry: 
   f(x) = x^3 + 1
   f(-x) = (-x)^3 + 1 = -x^3 + 1
   Since f(x) is not equal to f(-x), the function is not even.

2. Check for odd symmetry: 
   f(x) = x^3 + 1
   -f(-x) = -(-x)^3 - 1 = x^3 - 1
   Since f(x) is not equal to -f(-x), the function is not odd.

Therefore, the function f(x) = x^3 + 1 is neither even nor odd.