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Find the Taylor polynomials of the third degree approximating the function 𝑓(π‘₯) = √π‘₯ at the point π‘Ž = 9.

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Video Transcript

Find the Taylor polynomials of the third degree approximating the function 𝑓 of π‘₯ equals the square root of π‘₯ at the point π‘Ž equals nine.

Let’s start this question by writing out the general form for Taylor polynomials up to degree 𝑛. And let’s begin by substituting in the point where we’re centering our approximation, at π‘Ž equals nine. Also note that we’ve also been asked to find the Taylor polynomials up to the third degree. From here, we can see that we’re going to need to find 𝑓 of nine, 𝑓 prime of nine, 𝑓 double prime of nine, and 𝑓 triple prime of nine. To do this, we’re going to need to find 𝑓 prime of π‘₯, 𝑓 double prime of π‘₯, and 𝑓 triple prime of π‘₯.

Well, we know from the question that 𝑓 of π‘₯ equals the square root of π‘₯, which we know is the same as π‘₯ to the half power. Writing it this way is going to help us differentiate it to find the first derivative. Using the power rule, which tells us to multiply by the power and then subtract one from the power, gives us the first derivative. 𝑓 prime of π‘₯ equals one over two π‘₯ to the power of negative one-half.

And we then differentiate this to get the second derivative of 𝑓. Applying the power rule again gives us that the second derivative of 𝑓 is negative one over two multiplied by one over two π‘₯ to the power of negative three over two. But we can just simplify this to negative one over four π‘₯ to the power of negative three over two.

And we differentiate once more to get the third derivative of 𝑓. This gives us negative three over two multiplied by negative one over four π‘₯ to the power of negative five over two. But again, we can simplify this to give us three over eight π‘₯ to the power of negative five over two.

But what we actually want here is 𝑓 of nine, 𝑓 prime of nine, 𝑓 double prime of nine, and 𝑓 triple prime of nine. So we substitute nine into each of these functions. And then we substitute the values that we found back into our working. And from here, we just need to do some simplification to achieve our final answer.

If we start here with one over six over one factorial and we recall that 𝑛 factorial is the product of 𝑛 and all the integers below it down to one. Then one factorial is just one. So one over six over one factorial is just one over six over one, which is just one over six.

Now let’s simplify negative one over 108 divided by two factorial. Well, two factorial is just two multiplied by one, which is two. So this is just negative one over 108 over two. And we know from fraction laws that this is just negative one over 108 multiplied by two, which is negative one over 216. And finally, we need to simplify one over 648 divided by three factorial. Well, three factorial is three multiplied by two multiplied by one, which is six. So this is just one over 648 multiplied by six, which gives us one over 3888. So that gives us our final answer. The Taylor polynomials of the third degree approximating the function 𝑓 of π‘₯ equals the square root of π‘₯ at the point π‘Ž equals nine.

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