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F b N X { a No601. Oct 15, 12Question Show that the derivative of f(x) = (xa)m (xb)n vanishes at some point between a and b if m and n are positive integers My attempt f(x) = (xa)m (xb)n f '(x) = m(xa)m1 (xb)n n(xa)m (xb)n1 f '(x) = (xa)n1 (xb)n1 (m)(xb) (n)(x. A f B _ X @ t b g T p N N V E e RZ Ă ݂ O v @ ` P b g.

F(x) hasa value equal to f(c) = ∫b a f(x)dx b a Multiplying bothsides by b a proves the result 4The first fundamental theorem of integral calculus We are now in a position to prove our first major result about the definite integral The result concerns the socalled area function F(x) = ∫ x a f(t)dt and its derivative with respect to x. It is highly appreciated if you tell us what you tried, what you think about the solution, etc So since you didn't do that I'll give you a solution for only a piece of the question, and I hope you can complete by your own. I ~ C f b N X new.

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F b N X { a No605. N is also the coe cient on the highestdegree term, xn 1, in g(x), so g(x) is degree n 1, as desired Part 52 For each real a, the function p given by p(x) = f(x a) is a polynomial of degree n Solution We induct on n to prove this statement4 Base Case We could use n = 0 as the base case, but for clarity will show n = 1 When n = 1, f(x. ЃC G k f B b N X ̃z y W ł B.

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