F'(x) is an even function, state the correct answer:

State the correct answer:
f'(x) is an even function with domain all real numbers which is continuous for all values of x and differentiable for all values of x

A- f(x) is always an odd function
B- f(x) is not always an odd function

Justify your answer

Please Help!

Let f(x) = x + 1.
f(1) = 2
f(-1) = 0
f(-1) ≠ -f(1)
This is not an odd function.

f'(x) = 1
f'(-x) = 1
f'(-x) = f'(x)
This is an even function.

There exists at least one case in which f'(x) is and even function, but f(x) is not an odd function, so B is the correct answer.

(B) is correct.

For example f(x) = x^3 + 1 is not odd (example: f(-1) = (-1)^3 + 1 = -1 + 1 = 0, but f(1) = 1^3 + 1 = 2, so f(-1) isn't -f(1)) but f'(x) = 3x^2 is certainly even.

What you can say in general is that the function f(x) - f(0) must be an odd function. But f(x) itself need not be (as the above example shows). To see that g(x) = f(x) - f(0) is odd, use the fundamental theorem of calculus to write g(x) = f(x) - f(0) as the integral from 0 to x of f'(t) dt (you interpret this in the usual way, as -[the integral from x to 0 of f'(t) dt], when x is negative]. You can then see for any x that

g(-x) = f(-x) - f(0)
= the integral from 0 to -x of f'(t) dt
[let s = -t, then ds = -dt and the new bounds are -0 and -(-x) = x]
= the integral from 0 to x of f'(-s) (-1) ds
[use evenness of f']
= the integral from 0 to x of f'(s) (-1) ds
= - the integral from 0 to x of f'(s) ds
[use the fundamental theorem of calculus again]
= -[f(x) - f(0)]
= -g(x)

showing that g(x) is odd.