Consider this statement about a function f(x):
(*) If (f(x))2 ≤ 1 for all −1 ≤ x ≤ 1 then ∫−11(f(x))2 dx ≤ ∫−11 f(x) dx
Which one of the following functions provides a counterexample to (*)?
- Af(x) = x + 12
- Bf(x) = x − 12
- Cf(x) = x + x3
- Df(x) = x − x3
- Ef(x) = x2 + x4
- Ff(x) = x2 − x4
Show the answer and worked solution
answer · D
- Af(x) = x + 12
- Bf(x) = x − 12
- Cf(x) = x + x3
- Df(x) = x − x3
- Ef(x) = x2 + x4
- Ff(x) = x2 − x4
A counterexample has to satisfy the hypothesis and break the conclusion, so first throw out anything with (f(x))2 > 1 somewhere on [−1, 1]: at x = 1, option A gives f = 32, option C gives f = 2 and option E gives f = 2, while option B gives f(−1) = −32. That leaves D and F. For f(x) = x − x3 the largest value of |f| on [−1, 1] is 23√3 ≈ 0.385, so the hypothesis holds; but f is odd, so ∫−11 f dx = 0 while ∫−11 f2 dx > 0. The conclusion fails, so D is the counterexample. For F, 0 ≤ f ≤ 14, so f2 ≤ f and the conclusion holds.