Suppose that the function f : R → R is differentiable. Define the function H: R → R by
H(x) = integral from −x to x of [ f (t) + f (−t)]dt for all x in R.
Find H'' (x).
solution: i divided it up into int 0 to x of [ f (t) + f (−t)]dt and negative int from 0 to -x of [ f (t) + f (−t)]dt. then by the 2nd FTC, H '(x)=
2[ f (x) + f (−t)], so H ''(x) should then be of 2[ f '(x) - f '(−x)]. I'm just not sure if the f(-t) changes anything besides the fact that when I differentiate f(-t) I will use the chain rule and multiply by -1.
