# Video: Finding the Unknown Length in a Right Triangle Using Trigonometry Where the Unknown Is on the Bottom of the Fraction

Find the Taylor polynomials of degree two approximating the function π(π₯) = π₯Β³ + 2π₯ β 3 at the point π = 2.

02:07

### Video Transcript

Find the Taylor polynomials of degree two approximating the function π of π₯ equals π₯ cubed add two π₯ minus three at the point π equals two.

Letβs start by writing out the general form of a Taylor polynomial up to degree π for a function π. And for this question, weβre centering our approximation at the point π equals two. So we make the substitution π equals two. Also note that weβve only gone up to the degree two power as thatβs what weβve been asked to do in the question.

So now we need to evaluate π of two, π prime of two, and π double prime of two. Well, we know that π of π₯ equals π₯ cubed add two π₯ minus three. And so π of two equals two cubed add two multiplied by two minus three. So π of two equals nine.

To find π prime of two, we need to, first of all, find π prime of π₯. To do this, we differentiate π of π₯, which we do term by term using the power rule, to get π prime of π₯ equals three π₯ squared add two. Note that negative three differentiates to zero because itβs a constant. And so π prime of two equals three multiplied by two squared add two, which gives us 14.

To find π double prime of two, we need to, first of all, find π double prime of π₯, which we find by differentiating the first derivative, to get six π₯. And so π double prime of two equals six multiplied by two, which is 12.

We then substitute these values into our workings. From here, we can evaluate these factorials. We know that the factorial of a number is the product of that number and all the integers below it to one. So one factorial is one multiplied by one, which is one. And two factorial is two multiplied by one, which is two. From here, we just need to do some simplification. 14 over one is just 14, and 12 over two is just six. And so that gives us our final answer: nine add 14 multiplied by π₯ minus two add six multiplied by π₯ minus two squared.