Worked example: coefficient in Taylor polynomial | Series | AP Calculus BC | Khan Academy

Given an f of x, and they say, what is the coefficient for the term containing x plus 2 to the 4th power in the Taylor polynomial centered at x equals negative 2?

So, like always, take a see if you can take a stab at this video on your own before we work through it together.

All right, now let's do this.

In general, our Taylor polynomial P of x is going to have the form—and remember we're centering it at x equals negative two—so this means we're going to evaluate our function at where we're centering it.

We are going to divide it by zero factorial, which is just one. I'm just going to write it all out just so you see the pattern. We could even say that's going to be times x minus where we're centering it, but if we're subtracting a negative 2, it's going to be x plus 2.

I could write to the 0th power, but once again, that's just going to be 1. So a lot of times you won't see someone write this and this, but I'm writing it just to show that there's a consistent pattern.

So then you're going to have plus the first derivative evaluated at negative 2 divided by 1 factorial, which is still just 1, times x plus 2 to the first power plus the second derivative evaluated at negative 2 over 2 factorial times x plus 2 squared.

I think you see where this is going, and really all we care about is the one that has a fourth degree term.

Well, actually let me just write the third degree term too, just to get fluent in this. So the third derivative evaluated at negative two over three factorial times x plus two to the third power.

And now this is the part that we really care about: plus the fourth derivative— I could have just written a 4 there, but I think you get what I'm saying— and then evaluate at x equals negative 2 divided by 4 factorial times x plus two to the fourth power.

So what's the coefficient here? Well, the coefficient is this business. So we need to take the fourth derivative of our original function.

We need to take the fourth derivative of that original function evaluated at negative two and divided by four factorial.

So let's do that. So our function— our first derivative f prime of x is just going to be, just gonna use the power rule: 6 x to the fifth minus three x squared.

The second derivative is going to be equal to five times six is thirty x to the fourth, two times three minus six x to the first power.

The third derivative— the third derivative of x is going to be equal to 4 times 30, which is 120 x to the third power minus 6.

And then the fourth derivative, which is what we really care about, is going to be 3 times 120, which is 360 x to the second power, and the derivative of constants is just 0.

So if we were to evaluate this at x equals negative two, so f the fourth derivative evaluated when x equals negative two is going to be 360 times negative two squared, which is 4.

I'm just going to keep that as 360 times 4. We can obviously evaluate that, but we're going to divide it by 4 factorial.

So the whole coefficient is going to be 360 times 4, which is the numerator here, divided by 4 factorial, divided by 4 times 3 times 2 times 1.

Well, 4 divided by 4, that is going to be 1. 360 divided by 3— maybe I'll think of it this way: 360 divided by 6 is going to be 60, and so that's all we have.

We have 60, and then the denominators have a 1. So this is going to simplify to 60. That's the coefficient for this term.