Video: Find a Derivative Value from a Taylor Series

The fourth-degree Taylor polynomial of 𝑓 centered at π‘₯ = 2 is given by 𝑝(π‘₯) = 1 + (π‘₯ βˆ’ 2) + 5(π‘₯ βˆ’ 2)Β² + (7/3) (π‘₯ βˆ’ 2)Β³ + (3/4) (π‘₯ βˆ’ 2)⁴. What is 𝑓‴(2)?

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Video Transcript

The fourth-degree Taylor polynomial of 𝑓 centered at π‘₯ equals two is given by 𝑝 of π‘₯ equals one add π‘₯ minus two add five π‘₯ minus two squared add seven over three π‘₯ minus two cubed add three over four π‘₯ minus two to the fourth power. What is 𝑓 triple prime of two?

Firstly, recall that this notation means the third derivative of 𝑓 evaluated at two. This is the general form for Taylor polynomial of a function 𝑓 centered at π‘₯ equals π‘Ž. So if we center this general Taylor polynomial at π‘₯ equals two, this is what it becomes. Now, we can actually compare this with the Taylor polynomial in our question. We can see how the π‘₯ minus two, π‘₯ minus two squared, π‘₯ minus two cubed, and so on all correspond. So we can also see how the coefficients correspond.

Remember, we’re trying to find 𝑓 triple prime of two. So we can equate the coefficients of π‘₯ minus two cubed. That is, seven over three and 𝑓 triple prime of two over three factorial. Recall that the factorial of a number is the product of that number and all the integers below it to one. So three factorial is three times two times one, which is six.

We then want to rearrange to find 𝑓 triple prime of two. We can do this by multiplying both sides by six. So we have that six multiplied by seven over three equals 𝑓 triple prime of two. But six multiplied by seven over three is just 42 over three. But 42 over three just cancels down to 14. So we find that 𝑓 triple prime of two equals 14.

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