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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.

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