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# Use the definition of a Taylor series to find the first four nonzero terms of the series for $f(x)$ centered at the given value of \$ a. $$f(x) = \sqrt [3]{x}, \quad a = 8$$

## $$2+\frac{1}{12}(x-8)-\frac{1}{3^{2} \cdot 2^{5}}\left(x{-8}\right)^{2}+\frac{10}{3^{3}} \cdot \frac{1}{2^{5} \cdot 3 !}(x-8)^{3}$$

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the problem is, use the definition off a tiller. Siri's by the first of one answer returns over the Siri's for half our backs entered. That's given money off, eh? So first half it is equal to two. The first term it is, too. I have promise he could do one over read hams X to the power off negative to third. So have prom. At this point eight, it's equal to one hams, eight to the power of negative to third, which is equal to third terms. One over four, which is equal to one twelve. It's a second term is one over twelve times X one of ST Second derivative is a one off three times negative to over three, thanks to the public. Negative far over three. I have proof from from pate. It's the heart too negative. Two over nine times, one over already too, which is equal to negative one over three square taps through through the power for so the third term, this one over. Negative one over three. Squire task through to the parlour for you, Holmes to bacterial. How sax months take sport, which is the connective one over three squared, perhaps to into the power X minus. Paid scored Third Dimension as they called you one over three counts negative to over three. I'm selective while off. Three. House Extra negative. Eight over three After from from trumpet, it's a call to ten hour. Three skew comes one over two to the power of it. So close to term. This ten over three skew has won over you in pain. Three. Factorial ex st. It's a problem three.

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