(7pts) Explain why for any0?n?k?1,f(n+1)is bounded on[a, b], i.e. that there existsM >0such that|f(n+1)(s)| ?Mfor alls?[a, b].

yx

f?(s)ds. Show that 2. Assume the formula(1)is true for some0?n?k?2. Consider the functions:

this is true using the Fundamental Theorem of Calculus.

yx

2. (6pts) Prove of Calculus.

M(s?y)nds=?M(x?y)n+1/(n+1)using the fundamental Theorem(?1)ny|y?x|n+1

3. (7pts) Prove the inequalityn!(s?y)nf(n+1)(s)ds?M(n+ 1)!.x

Interpretation: in the formula (1), asyandxare closer and closer, we get an approx- imation off(y) by the polynomialf(x) + (y?x)f?(x) +...+ (y?x)nf(n)(x)!, and the remainder is "smaller" because in the inequality above|y?x|n+1decreases faster than|y?x|n.

1. ( 7pts ) Explain why for any O O such that I f(~+ 1 ) ( S ) |