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# Worked example: coefficient in Taylor polynomial

AP.CALC:
LIM‑8 (EU)
,
LIM‑8.A (LO)
,
LIM‑8.A.1 (EK)
,
LIM‑8.A.2 (EK)
,
LIM‑8.B (LO)
,
LIM‑8.B.1 (EK)

## Video transcript

give it an f of X and they say what is the coefficient for the term containing X plus 2 to the fourth power in the Taylor polynomial centered at x equals negative 2 of F so like always take us if you could take a stab at this video on your own before we work through it together all right now let's let's do this so in general our Taylor polynomial P of X it's going to have the form and remember we're centering it x equals negative 2 so this means we're going to evaluate our function at where we're centering it we are going to divide it by 0 factorial which is just 1 I'm just going to write them all out just so you see the pattern and we could even say that's going to be times X minus where we're centering it but if we're subtracting a negative 2 is going to be X plus 2 and I could write to the 0 power once again that's just going to be 1 so a lot of times you won't see someone write this and this but I'm writing it just to show that there's a consistent pattern so then you're going to have plus the first derivative evaluated at negative 2 divided by 1 factorial which is still just 1 times X plus 2 to the first power plus the second derivative evaluated at negative 2 over 2 factorial times X plus 2 squared I think you see where this is going and really all we care about is the one that has the fourth degree term and I will actually just write the third degree term to just we get fluent doing this so the third derivative evaluated at negative 2 over 3 factorial times X plus 2 to the third power and now this is the part that we really care about plus the fourth derivative I could have just written a 4 there but I think you get what I'm saying and then evaluate at x equals negative 2 divided by 4 factorial times X plus 2 to the fourth power so what's the coefficient here well the coefficient is this business so we have take the fourth derivative of our original function we take the fourth derivative that original function evaluated at negative 2 and divided by 4 factorial so let's do that so our function so our first derivative F prime of X is just going to be just going to use the power rule a lot 6x to the fifth minus 3x squared second derivative is going to be equal to 5 times 6 is 30 X to the fourth 2 times 3 minus 6 X to the first power third derivative third derivative of X is going to be equal to 4 times 30 is 120 X to the third power minus 6 and then the fourth derivative which is what we really care about is going to be 3 times 120 is 360 X to the second power and the derivative of constants just zero so if we were to evaluate this at x equals negative 2 so f the fourth derivative evaluated when x equals negative 2 is going to be 360 times negative 2 squared is 4 I'm just going to keep that as 360 times 4 we can obviously evaluate that but we're going to divide it by 4 factorial so the whole coefficient is going to be 360 times 4 which is the numerator here divided by 4 factorial divided by 4 times 3 times 2 times 1 well 4 divided by 4 those is going to be 1 360 divided by 3 maybe I'll think of it this way 360 divided by 6 is going to be 60 and so that's all we have we have 60 and then the denominators have a 1 so this is going to simplify to 60 that's the coefficient for this term
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