Find the Taylor polynomials of
degree two approximating the function 𝑓 of 𝑥 equals 𝑥 cubed add two 𝑥 minus
three at the point 𝑎 equals two.
Let’s start by writing out the
general form of a Taylor polynomial up to degree 𝑛 for a function 𝑓. And for this question, we’re
centering our approximation at the point 𝑎 equals two. So we make the substitution 𝑎
equals two. Also note that we’ve only gone up
to the degree two power as that’s what we’ve been asked to do in the question.
So now we need to evaluate 𝑓 of
two, 𝑓 prime of two, and 𝑓 double prime of two. Well, we know that 𝑓 of 𝑥 equals
𝑥 cubed add two 𝑥 minus three. And so 𝑓 of two equals two cubed
add two multiplied by two minus three. So 𝑓 of two equals nine.
To find 𝑓 prime of two, we need
to, first of all, find 𝑓 prime of 𝑥. To do this, we differentiate 𝑓 of
𝑥, which we do term by term using the power rule, to get 𝑓 prime of 𝑥 equals
three 𝑥 squared add two. Note that negative three
differentiates to zero because it’s a constant. And so 𝑓 prime of two equals three
multiplied by two squared add two, which gives us 14.
To find 𝑓 double prime of two, we
need to, first of all, find 𝑓 double prime of 𝑥, which we find by differentiating
the first derivative, to get six 𝑥. And so 𝑓 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 𝑥 minus two add six multiplied by 𝑥 minus two