We want to find the first few Taylor Polynomials
of the given function centered at a equal zero.
So for a quick review,
a Taylor Polynomial is formed
when we use a finite number of terms
from a Taylor series given here,
where a Taylor series is a representation of a function
as an infinite series calculated from
the values of its derivatives at a single point.
If f of x is continuously differentiable at x equals c,
then the Taylor series is defined here
when the numerator is always going to be the nth derivative
valuated at c, divided by n-factorial,
times the quantity x minus c, raised to the power of n.
Notice how the first two terms
don't have a denominator,
and that's just because
when n is zero, we'd have zero-factorial,
the denominator, which is one.
And then when it is one, we'd have one-factorial
the denominator, which is also one.
But in our case, because
we want the polynomial to be centered at x equals 0,
we can call the Taylor series, or Maclaurin series,
and when we form our polynomial
using a finite number of terms from the Maclaurin series,
we can call it a Maclaurin polynomial.
Notice here, we're always evaluating
the derivatives at x equals zero,
and instead of having the quantity x minus c,
raised to the nth,
we just have x to the nth,
because again, it's centered at zero.
So going back to our example,
because we're only looking for the degree one
Taylor polynomial,
or if we want, Maclaurin polynomial,
and the degree two Maclaurin polynomial,
we only have to find f-prime of zero,
and f-double prime of zero,
before we apply our formula.
We'd begin by finding the first and second derivative though
of the given function.
So looking at our function, that is f-prime of x...
Would be equal to the derivative of 5e to the x,
which 5e to the x,
plus the derivative of 3e to the -x,
which would be 3e to the -x times -1,
so we'd have -3e to the -x.
Let's also find the second derivative.
The second derivative is a derivative
of the first derivative,
so we'd have 5e to the x,
and -3e to the -x times -1,
so we'll have +3e to the -x.
Notice how the second derivative
is the same as the original function.
So f of zero would be 5e to the zero
plus 3e to the zero,
which would be five plus three, or eight,
and now f-prime of zero would be
5e to the zero minus 3e to the zero,
which would be five minus three, or two,
and then f-double prime of zero,
is going to be the same as f of zero.
We'd have 5e to the zero,
plus three to the zero,
which again is eight.
Instead of finding the degree one Taylor polynomial,
identified here by T sub one of x,
we'll only be using the first two terms
of the Taylor series, or if we want, Maclaurin series.
So we would have f of zero, which is eight
plus f-prime of zero times x, which would be 2x.
Notice how this is a linear approximation
of the given function of x equals zero,
which would just be the tangent line at x equals zero.
And now for the degree two Taylor polynomial.
Notice how we'll use the first three terms
of the Taylor or Maclaurin polynomial.
So the first two terms would be the same,
we'd have eight plus 2x,
then we'll have plus f-double prime of zero,
which is eight,
divided by two-factorial, times x-squared.
Well, two-factorial's equal to two times one or two,
so this simplifies to eight plus 2x,
plus 4x-squared.
Now take a look at the graph of the original function
under degree one and degree two
polynomial approximations for the function
at x equals zero.
So in blue we have the graph of the original function.
In red we have the degree one Taylor polynomial,
which again we can also call the Maclaurin polynomial,
because it is centered at x equals zero.
And then in green, we have the
degree two Taylor polynomial.
So as you can imagine,
if we were to add more terms to our Taylor polynomial
or Maclaurin polynomial,
the approximation would get better and better.
I hope you found this helpful.