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Evaluate the given functions by using three terms of the appropriate Taylor series.$$\sqrt{9.3}$$

Calculus 2 / BC

Chapter 30

Expansion of Functions in Series

Section 5

Taylor Series

Series

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Okay. So for this problem we are asked to approximate squared of 9.3 by using taylor series expansion. So our function is the square root of X, where X is equal to 9.3. So we're gonna numerically approximate 9.3 by using uh squared of 9.3. Sorry, by using this uh expression for taylor series expansion. So we have this uh throughout this formula, there's lots of values for A. So what is A? So the value of A must be nearest to X. Such that X minus A should be smaller. So since X is 9.3 will let a equal nine. Okay, so let's get started. We know that our function 1/2 is equal to square root of X. So we need to find F evaluated at A. So F of A is equal to three. Okay, now we need to find our next piece so we need to find uh the derivative of X. So that will just be 1/2 x To the 1/2 power. Okay, now we need to evaluate are derivative at A And that will just give us 1/6. All right. So now we need to find our second derivative and evaluate it at A. So f double prime of X is equal to So of -1 over for acts to the three over to power. Okay, So now we're going to Evaluate our second derivative at a and that will just give us -1 over 108. Okay. We have everything that we need. So therefore we can say at square root of 9.3 is equal to school of three Plus 1/6. 9.3 -9 Will have -1 over to 16. The reason why is because we have a Two factorial here over 108 here. So we're just gonna multiply that by two that will give us uh, went over to 16. So then we'll have 9.3 -9 squared plus additional terms. So after throwing this in your calculator, we'll get that square root of 9.3 is equal to three zero for 96

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