### Video Transcript

Find the error bound π
three when using the third Maclaurin polynomial for the function π of π₯ is equal to π to the power of π₯ at π₯ is equal to zero to approximate the value of π to the power of negative 0.2.

To answer this question, we recall Taylorβs theorem, which tells us that if a function π and its π plus one derivatives, π one up to ππ plus one, are continuous on an interval containing π and π₯. Then we can conclude that π of π₯ is equal to the π term Taylor polynomial, which we call ππ of π₯. So we add to this the remainder term, which we call π
π of π₯. To add to this, we know that the absolute value of our remainder term at π₯ is less than or equal to the absolute value of some constant π multiplied by π₯ minus π all raised to the power of π plus one divided by π plus one factorial.

And we can think of this constant π as being an upper bound on the absolute value of our π plus one derivative function on some interval containing both π₯ and π. Since the question wants us to find the error bound for π
three. We can conclude that we must want π equal to three. Next, the question asks us to use the third Maclaurin polynomial for the function π of π₯ is equal to π to the power of π₯. And we know that the Maclaurin polynomial of a function is a special example of the Taylor polynomial of a function. Itβs when we set π equal to zero. The question asks us to approximate the function π of π₯ is equal to π to the power of π₯. So that set our function π equal to π to the π₯.

Finally, the question wants us to approximate the value of π to the power of negative 0.2. So we will set π₯ equal to negative 0.2. The first thing weβre going to want to do is attempt to find a value for π an upper bound on the absolute value of the π plus one derivative of a function π on some interval containing both π₯ and π. Well, we remember that the question tells us weβre looking for third-degree approximation. So π is equal to three and we want the fourth derivative of a function π of π₯. So letβs now try to find the fourth derivative of a function π of π₯.

Well, the first derivative π one of π₯ is equal to the derivative with respect π₯ of π to the π₯. And we know that this derivative is just equal to π to the power of π₯. In fact, every time we take the derivative with respect to π₯ of π to the power of π₯, weβll still just get π to the power of π₯. So our fourth derivative function is still going to be π to the power of π₯. Now that we have our fourth derivative function, we need to find an upper bound on the absolute value where π₯ is between negative 0.2 and zero. Letβs start by writing the absolute value of our fourth derivative function thatβs equal to the absolute value of π to the power of π₯. In fact, we know the π to the power of π₯ is positive. So the absolute value of π to the power of π₯ is just equal to π to the power of π₯.

We want to know what the largest possible output for a function π to the power of π₯ is when π₯ is between negative 0.2 and zero. Well, since our function is π to the power of π₯, we want π₯ to be as large as possible. That will give us the large as possible output. So this is when π₯ is equal to zero. And since π to the zeroth power is just equal to one, we must have the π to the π₯ is less than or equal to one when negative 0.2 is less than or equal to π₯ is less than or equal to zero. So weβre going to set the value of π equal to one. Weβre now ready to find the error abound when using the third Maclaurin polynomial for the function π of π₯ is equal to π to the power of π₯. Since the question asked us to find the error bound π
three, thatβs the third-degree approximation, we substitute π is equal to three.

Next, since weβre asked to approximate π to the power of negative 0.2, we want to approximate the value of our function when π₯ is equal to negative 0.2. So letβs substitute in π₯ is equal to negative 0.2. Since we want to use the Maclaurin polynomial, we know that this is a special case when we set π equal to zero. Finally, we showed earlier that we could set π equal to one. This gives us that the absolute value of our error bound π
three is less than or equal to the absolute value of one multiplied by negative 0.2 minus zero all raised to the power of three plus one all divided by three plus one factorial.

If we then evaluate this, we get the absolute value of negative 0.2 raised to the fourth power divided by four factorial. Finally, we can evaluate this expression to give us that the error bound, π
three, when using the third Maclaurin polynomial for the function π of π₯ is equal to π to the power of π₯ at π₯ is equal to zero when approximating the value of π to the power of negative 0.2 gives us one divided by 15000.