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Wouldn't f(x)= (x+1)^3 be an odd function that is not true for the above statement?
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It would be odd under x+1 —> -(x+1), but then the integral is shifted by one unit and hence you lose the symmetry that makes the integral 0
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Falls under neither odd nor even?
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Yea it’s neither
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^^^ you may be looking at the degree. you can *verify* its parity via the definition of even and odd functions
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remember that, algebraically, a function is odd iff ∀ x ∊ D , f(x) = -f(-x) let x ∊ ℝ let f(x) = (x+1)^3 choose x = 2 f(2) = [(2)+1]^3 = (3)^3 = 27 -f(-2) = - [(-2)+1]^3 = - [-1]^3 = - [-1] = 1 27≠ 1, so f(2) ≠ -f(-2) i.e f(x) ≠ -f(-x) Hence, the function is (not) odd.
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