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**clockingly** g(x) = sin(x) + cos(.5x) on (-4, 4)

Okay, so for the above expression, I have to:

a) Calculate an expression for the 5th derivative g^(5)(x).

b) Find a bound for the absolute value of g^(5)(x) on the interval indicated.

c) Use Taylor's Theorem to find a cap for the error; that is, a cap for the difference between the function and the Maclaurin polynomial of degree four on the interval.

I did part a and this is what I got:

g'(x) = cos(x)-.5sin(.5x)

g''(x) = -.25cos(.5x) - sin(x)

g'''(x) = .125sin(.5x) - cos(x)

g^(4)(x) = .0625cos(.5x) + sin(x)

g^(5)(x) = cos(x) - .03125sin(.5x)

The last one is the 5th derivative.

I attempted part b and basically what I did was substitute 4 for x in the expression for the 5th derivative of g(x). From this I got approximately .68, which I rounde up to .7. Then for a final answer, I got

the absolute value of cos(x) - .03125sin(.5x) <= .7

I'm not sure if the above is right, but that's what I got.

My real problem lies in doing part c. I guess I'm confused as to the method because my book isn't very clear.