### Video Transcript

Estimate the function π of π₯ is
equal to two times the sin of three π₯ with a third-degree Taylor polynomial at π₯
is equal to π.

The question is asking us to find a
Taylor polynomial approximation of third degree at π₯ is equal to π of the function
π of π₯ is equal to two times the sin of three π₯. We recall an πth-degree Taylor
polynomial approximation of the function π of π₯ at π₯ is equal to π is given by
π π of π₯ is equal to π evaluated at π. Plus the first derivative of π
evaluated at π divided by one factorial multiplied by π₯ minus π. Plus the second derivative of π
evaluate at π divided by two factorial multiplied by π₯ minus π squared.

And we keep adding terms of this
form all the way up to the πth derivative of π evaluated at π divided by π
factorial multiplied by π₯ minus π to the power of π. We call π the degree of our Taylor
polynomial because it is also the highest power of π₯ which appears in the Taylor
polynomial. We call π the center of our Taylor
polynomial.

Weβre asked to find the
third-degree Taylor polynomial of our function. This means weβll need to find the
third derivative of our function π of π₯. So, weβll set π equal to
three. And since we want to find our
polynomial at π₯ is equal to π, weβll set our center π equal to π. To find the third-degree Taylor
polynomial approximation of our function, we need to find the first, second, and
third derivative of our function π of π₯.

Since we need to evaluate this at
π₯ is equal to our center π, weβll start by finding the first derivative of π of
π₯. To differentiate π of π₯, which is
two times the sin of three π₯, we recall, for constants π and π, the derivative of
π sin of ππ₯ with respect to π₯ is equal to ππ times the cos of ππ₯. Applying this, we get π prime of
π₯ is equal to two times three times the cos of three π₯. And we can simplify two times three
to just be six.

Now, to find the second derivative
of our function π of π₯, we need to differentiate the first derivative of our
function π of π₯. Thatβs the derivative of six times
the cos of three π₯ with respect to π₯. And to differentiate this, we
recall, for constants π and π, the derivative of π cos of ππ₯ with respect to π₯
is equal to negative ππ times the sin of ππ₯. Applying this to our function six
times the cos of three π₯ gives us our second derivative of π of π₯ is equal to
negative six times three times the sin of three π₯. And we can simplify negative six
times three to be negative 18.

To find our third derivative of π
of π₯, we need to differentiate our second derivative of π of π₯. Thatβs the derivative of negative
18 times the sin of three π₯ with respect to π₯. Using our first rule for
differentiating trigonometric functions, we have that this is equal to negative 18
times three times the cos of three π₯. And we have negative 18 times three
can be simplified to negative 54.

We see, in our Taylor polynomial
expansion, we want to evaluate each of our derivative functions at the value of the
center π. In our case, the center of our
Taylor approximation is π. So, letβs evaluate π, π prime, π
double prime, and π triple prime at π₯ is equal to π. We have that π evaluated at π is
equal to two times the sin of three π. However, the sin of three π is
just equal to zero, so π evaluated at π is equal to zero.

Next, we have π prime evaluated at
π₯ is equal to π is equal to six times the cos of three π. The cos of three π is equal to
negative one. So, this evaluates to give us
negative six. Our second derivative function
evaluated at π₯ is equal to π gives us negative 18 multiplied by the sin of three
π. We know the sin of three π is
zero, so this evaluates to give us zero.

Finally, our third derivative of π
of π₯ evaluated at π₯ is equal to π gives us negative 54 times the cos of three
π. And the cos of three π is negative
one. So, this evaluates to give us
54. Weβre now ready to find our
third-degree Taylor approximation of two times the sin of three π₯ at π₯ is equal to
π. Itβs equal to π of π. Plus π prime of π over one
factorial multiplied by π₯ minus π. Plus π double prime of π over two
factorial times π₯ minus π squared. Plus π triple prime of π over
three factorial times π₯ minus π cubed.

We already found the values of π
evaluated at π, π prime evaluated at π, π double prime evaluated at π, and π
triple prime evaluated at π. Substituting these values into our
Taylor approximation, we see that the first term and the third term have a factor of
zero. So, theyβre equal to zero. So, this simplifies to give
negative six over one factorial multiplied by π₯ minus π plus 54 over three
factorial multiplied by π₯ minus π cubed.

One factorial is just equal to
one. So, negative six divided by one
factorial is negative six. And three factorial is equal to
six. So, 54 divided by three factorial
is equal to nine. Therefore, weβve shown that the
~~first-degree~~ [third-degree] Taylor polynomial approximation at π₯ is
equal to π of the function π of π₯ is equal to two times the sin of three π₯ is
equal to negative six times π₯ minus π plus nine times π₯ minus π cubed.