Calculus II: Taylor's inequality

[@MathforThought]

At 18:25 I wrote R_3(3) instead of R_2(3). That was a typo.

[@nicolasazzolini2218]

Would that mean the factorial is 3! and |x-4|^3 rather than the ones in the video?

[@MathforThought]

@@nicolasazzolini2218 Correct. The answer is still right though.

[@BrunoSilva-oz5es]

why choose such a hard deritive for the first problem?

[@pear3663]

For example 2; when finding the third term of the Taylor series to use in the A.S.E.T., why do we neglect the (x-4)^3 part of the third term and just put f'''(4)/3! for b_3?

[@MathforThought]

We just care about the coefficent.

[@artart421]

For example 2, shouldn't it be |R2(5)|

[@MathforThought]

No. You need the largest possible value of M. That occurs at 3. Not 5.

[@artart421]

For the ASET on example 2, why did we go with f^4(4)?

[@MathforThought]

But I didn't... I did f'''(4).

[@harshkothari5738]

So to calculate the highest value for M we try to find critical numbers for that and also evaluate at the ends of the intervals but my question is that if the critical point is not in the interval shouldn't we ignore it anyway?

[@MathforThought]

It should be included. Otherwise it makes so sense to check the interval.

[@harshkothari5738]

@@MathforThought so first we check the critical point and if it is continuous on there, we use that even if it isn't in the interval correct?

[@MathforThought]

@@harshkothari5738 Sorry can you clarify? If the critical point isn't in the given interval, you don't use that.

[@harshkothari5738]

@@MathforThought nevermind I understand now