Suppose f is a function with the property that abs(f(x)) <= x^2 for all x. Show that f(0)=0. Then show that f'(0)=0.
Explanation / Answer
For any x we are given that x^2>=|f(x)| and |f(x)|>=0 by definition. Therefore, x^2 >= |f(x)| >= 0 At x=0, x^2=0 so 0 >= |f(0)| >= 0 and therefore |f(0)| = 0 --> f(0) = 0 ===== Let g(x) = x^2 then g'(x) = 2x and g'(0) = 0 Let h(x) = g(x) - |f(x)| = x^2 - |f(x)| h(0) = g(0) - |f(0)| = 0 - 0 = 0 For any arbitrarily small d>0 h(d) = g(d) - |f(d)| = d^2 - |f(d)| > 0 h(-d) = g(-d) - |f(-d)| = d^2 - |f(-d)| > 0 lim h(d) = h(0) so therefore h(0) is a critical point of h and h'(0) = 0 since g'(0)=0 f'(0) = 0 as well. === Edit: Yes I assumed |f(x)|