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Question Video: Use the Second-Degree Taylor Polynomial to Approximate a Function Mathematics • Higher Education

Use the second-degree Taylor polynomial to approximate the function 𝑓(π‘₯) = 2 + π‘₯ + π‘₯Β² at the point π‘Ž = 1.

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Video Transcript

Use the second-degree Taylor polynomial to approximate the function 𝑓 of π‘₯ equals two add π‘₯ add π‘₯ squared at the point π‘Ž equals one.

Let’s first of all remind ourselves of the general form of a Taylor polynomial. That is, we can write a function 𝑓 of π‘₯ as 𝑓 of π‘Ž add 𝑓 prime of π‘Ž over one factorial multiplied by π‘₯ minus π‘Ž add 𝑓 double prime of π‘Ž over two factorial multiplied by π‘₯ minus π‘Ž squared and so on. We’ve only been asked to go up to the second degree. So, we only need these first three terms.

In order to use this formula, we’re going to need 𝑓 of π‘Ž, 𝑓 prime of π‘Ž, and 𝑓 double prime of π‘Ž. To do this, we need to start by finding 𝑓 prime of π‘₯ and 𝑓 double prime of π‘₯. 𝑓 prime of π‘₯ is the derivative of 𝑓 of π‘₯ with respect to π‘₯. We can do this using the power rule, which tells us that the derivative with respect to π‘₯ of π‘₯ to the 𝑛th power is equal to 𝑛 multiplied by π‘₯ to the power of 𝑛 minus one. And we recall that constants differentiate to zero. So, two add π‘₯ add π‘₯ squared differentiates to give us one add two π‘₯.

We then differentiate the first derivative to get the second derivative. Again, remembering that constants differentiate to zero and using the power rule, we find that the second derivative is just two. But what we actually need for our formula is 𝑓 of π‘Ž, 𝑓 prime of π‘Ž, and 𝑓 double prime of π‘Ž. From the question, we’re told that π‘Ž equals one. This is where we’re centering our approximation. So, what we need to find is 𝑓 of one, 𝑓 prime of one, and 𝑓 double prime of one. We do this by substituting π‘₯ equals one into each function. And this gives us four, three, and two.

Returning to our formula, we can now substitute in what we found. So now that we’ve made substitutions for 𝑓 and its derivatives at the point one and π‘Ž equals one, we just need to simplify our answer. One factorial is just one times one. So, three over one factorial is just three. Two factorial is two multiplied by one, which is two. So, two over two factorial is just two over two, which is one. So, this leads us to our final answer: 𝑓 of π‘₯ is approximately four add three multiplied by π‘₯ minus one add π‘₯ minus one squared.

This is actually an interesting result, as if we expand our approximation, we actually get the polynomial we started with. But why is this the case? Well, we know the higher the degree we go up to with our Taylor polynomial, the better our approximation is. But if we expand our Taylor polynomial up to the same degree or higher than the degree of our original polynomial, we just get the original polynomial.

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