SOLUTION: Which of the following is NOT an odd function? A) f(x)=3x^3 + 5 B) f(x)=x C) f(x)=cos(x+pi/2) D) all of the above are odd functions

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Question 1199004: Which of the following is NOT an odd function?
A) f(x)=3x^3 + 5
B) f(x)=x
C) f(x)=cos(x+pi/2)
D) all of the above are odd functions

Found 2 solutions by math_tutor2020, Edwin McCravy:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Think of 5 as 5*1 = 5*x^0

The equation for choice (A)
f(x) = 3x^3 + 5
is equivalent to
f(x) = 3x^3 + 5x^0

We have an odd exponent 3 and an even exponent 0.
This polynomial function is neither even nor odd.
Even polynomial functions must have all exponents even.
Odd polynomial functions must have all exponents odd.

Choice (B) is an odd function since x = x^1 has an odd exponent.

Choice (C) appears to be an even function since cos(x) is even
But we have a phase shift of pi/2 units to the left, which means we really have a sine function here. Therefore, the function in choice (C) is odd.

You can use the identity
cos(A+B) = cos(A)cos(B)-sin(A)sin(B)
to find that
cos(x+pi/2) = -sin(x)

One common way to prove a function f(x) is odd is to show the following
f(-x) = -f(x)

Or you could use a visual approach to note how we have symmetry about the origin.

Any point (x,y) on the green curve has a corresponding mirror point at (-x,-y).

--------------------------------------------------

Answer: Choice (A) is not an odd function.



Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
A function is even when f(-x) = f(x) 
A function is odd when f(-x) = -f(x)

A)  
 <--not an odd function

B)  
 <--odd function

C) 
When you add 90o to an angle you get the sine in the NEXT quadrant,
which always has the opposite sign.
So 
     <--odd function

D) all the above are odd functions  <--false, A is not
Edwin
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