Video: Approximate a Function Using Taylor Polynomials

Find the Taylor polynomials of the fourth degree approximating the function 𝑓(π‘₯) = 𝑒^(2π‘₯) at the point π‘Ž = 3.

02:54

Video Transcript

Find the Taylor polynomials of the fourth degree approximating the function 𝑓 of π‘₯ equals 𝑒 to the two π‘₯ power at the point π‘Ž equals three.

Let’s start by writing out the formula for Taylor expansion. For this formula, π‘Ž is the value that we approximate the function at. So for this question, π‘Ž equals three. We’ve only been asked to go up to the fourth degree. So using the formula for Taylor expansion up to the fourth degree and substituting π‘Ž equals three gives us the Taylor expansion we’re going to work with. Now, there’s quite a few things we need to work out here. We’re going to need 𝑓 of three, the first derivative of 𝑓 at three, the second derivative of 𝑓 at three, the third derivative of 𝑓 at three, and the fourth derivative of 𝑓 at three.

So to do that, we’re going to need to find the values of the first, second, third, and fourth derivatives of 𝑓. So let’s calculate those. We have that 𝑓 of π‘₯ equals 𝑒 to the two π‘₯ power. And we differentiate this to get the first derivative. To do this, we recall that the derivative of 𝑒 to a function of π‘₯, 𝑔 of π‘₯ power, is equal to the first derivative of 𝑔 of π‘₯ multiplied by 𝑒 to the 𝑔 π‘₯ power. And as two π‘₯ differentiates to two, 𝑒 to the two π‘₯ power differentiates to two 𝑒 to the power two π‘₯. Differentiating two 𝑒 to the power two π‘₯ using the same rule gives us the second derivative. And differentiating again gives us the third derivative. And differentiating one more time gives us the fourth derivative. But what we actually need for our formula are these functions evaluated at π‘₯ equals three.

So we need to substitute three into each of these that we’ve just worked out. 𝑓 of three is 𝑒 to the power of two multiplied by three, which is 𝑒 to the sixth power. The first derivative evaluated at three is two 𝑒 to the power of two multiplied by three, which is two 𝑒 to the sixth power. And we do the same for the second, third, and fourth derivatives. So now I’m going to clear some space and substitute these values into our workings.

So from here, we can work out these factorials. We recall that the factorial of a number is the product of that number and all the integers below it up to one. So we can calculate these factorials to be one, two, six, and 24. From here, we just need to simplify some of these terms to get our final answer. Two 𝑒 to the sixth power over one is just two 𝑒 to the sixth power. Four 𝑒 to the sixth power over two is just two 𝑒 to the sixth power. Eight over six simplifies to four over three. And 16 over 24 simplifies to two over three. And so that gives us our final answer. The Taylor polynomials up to the fourth degree of the function 𝑓 of π‘₯ equals 𝑒 to the two π‘₯ power at the point π‘Ž equals three.

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