Video: Finding an Error Bound for a Taylor Polynomial Approximation of the Cube Root Function

Find the error bound when using the second Taylor polynomial for the function 𝑓(π‘₯) = π‘₯^(1/3) at π‘₯ = 8 to approximate the value of 𝑓(8.5). Give your answer in scientific form to three significant figures.

07:50

Video Transcript

Find the error bound when using the second Taylor polynomial for the function 𝑓 of π‘₯ is equal to π‘₯ to the power of one over three at π‘₯ equals eight to approximate the value of 𝑓 evaluated at 8.5. Give your answer in scientific form to three significant figures.

In this question, we’re trying to approximate the value of 𝑓 evaluated at 8.5 where 𝑓 of π‘₯ is the cube root of π‘₯ by using the second Taylor polynomial centered at the value of π‘₯ is equal to eight. We want to know how accurate this approximation is going to be. And to do this, we need to find an error bound for this approximation. And the question wants us to give this error bound in scientific form to three significant figures. Let’s start by recalling how we would find the error bound for an 𝑛th term Taylor polynomial approximation of a function 𝑓 of π‘₯.

We recall if 𝑓 of π‘₯ is equal to its 𝑛th term Taylor polynomial plus the remainder term 𝑅 𝑛 of π‘₯, then we can find an error bound for this remainder term. We know the absolute value of our remainder term 𝑅 𝑛, sometimes written 𝑅 𝑛 of π‘₯, will be less than or equal to the absolute value of 𝑀 times π‘₯ minus π‘Ž all raised to the power of 𝑛 plus one divided by 𝑛 plus one factorial, where 𝑀 is an upper bound on the absolute value of the 𝑛 plus one derivative of 𝑓 of π‘₯ on an interval which contains both π‘₯ and π‘Ž. And our error bound looks very complicated. However, we can make it simpler by just substituting the values given to us in the question. So we’ll start by doing this.

First, the question wants us to approximate our value by using the second Taylor polynomial, so we’ll set our value of 𝑛 equal to two. We can then update our error bound so that the value of 𝑛 is set to be two. Next, the question wants us to center our Taylor polynomial at π‘₯ is equal to eight. And our Taylor polynomial is centered at π‘Ž, so we’ll set the value of π‘Ž equal to eight. And of course, we’ll update our error bound so that the value of π‘Ž is set to be eight. Finally, the question wants us to approximate the value of 𝑓 evaluated at 8.5, so we should set our value of π‘₯ equal to 8.5. And just as we did before, we’ll update our error bound so that the value of π‘₯ is set to be 8.5.

There’s a couple of things worth pointing out here. First, we don’t set the value of π‘₯ in our bound for 𝑀 equal to 8.5. This is because 𝑀 is an upper bound on the absolute value of the 𝑛 plus oneth derivative of 𝑓 of π‘₯ on an interval containing both π‘Ž and π‘₯. This means the value of π‘₯ here is the variable and not the value of π‘₯ which we’re approximating in the question.

And it’s also worth pointing out we can’t do this for all functions. For example, we find the 𝑛 plus one derivative of 𝑓 of π‘₯. So we would need our function to be 𝑛 plus one times differentiable to even do this. In fact, there are more conditions to deal with this. For example, how many times can we differentiate 𝑓 of π‘₯ and are these derivatives continuous? However, we’re going to ignore this in this case because the question implies that we can do this anyway.

Now that we’ve substituted these values into our error bound, let’s take a closer look at our error bound. We can see the only thing we don’t know in our error bound is the value of 𝑀. So what we need to do now is find our value of 𝑀. We can see it’s an upper bound on the absolute value of the third derivative of 𝑓 of π‘₯ on an interval containing 8.5 and eight. We’ll choose the closed interval from eight to 8.5. This is because this is the smallest interval containing both of these values. And the smaller the interval we choose, the smaller our error bound will be.

Now, to find our value of 𝑀, we’re going to need to find an expression for the third derivative of 𝑓 of π‘₯ with respect to π‘₯. We’ll start by finding an expression for the first derivative of 𝑓 of π‘₯ with respect to π‘₯. That’s the derivative of π‘₯ to the power of one-third with respect to π‘₯. And we can do this by using the power rule for differentiation. We want to multiply by our exponent of π‘₯, which is one-third, and then reduce this exponent by one. We get 𝑓 prime of π‘₯ is equal to one-third times π‘₯ to the power of negative two over three.

We can do exactly the same to find the second derivative of 𝑓 of π‘₯ with respect to π‘₯. We multiply by our exponent of π‘₯, which is negative two-thirds, and then reduce this exponent by one to get negative five over three. This gives us 𝑓 double prime of π‘₯ is equal to one-third times negative two-thirds multiplied by π‘₯ to the power of negative five over three.

And we could simplify this expression. However, we only need to find an expression for the third derivative of 𝑓 of π‘₯ with respect to π‘₯, so this is not necessary. We’ll use the power rule for differentiation one more time to find an expression for 𝑓 triple prime of π‘₯. We get one-third times negative two-thirds multiplied by negative five over three times π‘₯ to the power of negative eight over three. And we can then simplify this expression. We get 𝑓 triple prime of π‘₯ is equal to 10 over 27 multiplied by π‘₯ to the power of negative eight over three.

We’re now ready to use this to find our value of 𝑀. First, we need to take the absolute value of the third derivative of 𝑓 of π‘₯. So we take the absolute value of both sides of this equation. We get the absolute value of the third derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to the absolute value of 10 over 27 multiplied by π‘₯ to the power of negative eight over three. And we can then simplify this expression. First, we see that 10 over 27 is a positive constant. Next, we need to recall we’re using this to find our value of 𝑀, which means we need to maximize this expression on the closed interval from eight to 8.5. This tells that our values of π‘₯ will be on the closed interval from eight to 8.5.

In particular, this means all of our inputs π‘₯ will be positive. And if π‘₯ is positive, π‘₯ to the power of negative eight over three is also positive. So this expression is always the product of two positive numbers. This means when we take its absolute value, we’re not changing its value. So we can just remove this symbol. So all we need to do is find an upper bound for this expression on the closed interval from eight to 8.5. To do this, we’ll use our laws of exponents. We’ll write π‘₯ to the power of negative eight over three in our denominator as π‘₯ to the power of eight over three. Doing this, we get the absolute value of the third derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to 10 divided by 27 times π‘₯ to the power of eight over three.

We want to find an upper bound for this expression on our interval, and we notice something interesting. 10 is positive, 27 is positive, and π‘₯ to the power of eight over three is positive on this interval. Therefore, to make this number as big as possible, we need to divide by the smallest positive number. So we need to choose the value of π‘₯ which makes π‘₯ to the power of eight over three as small as possible. And on our interval, this will be the smallest value of π‘₯, which is π‘₯ is equal to eight. Therefore, for π‘₯ in the closed interval from eight to 8.5, we’ve shown the absolute value of the third derivative of 𝑓 of π‘₯ with respect to π‘₯ is less than or equal to 10 divided by 27 multiplied by eight to the power of eight over three.

And we can evaluate this expression. If we do this, we get five divided by 3456. And of course, this will be our value of 𝑀. But now that we found our value of 𝑀, we found every single variable inside of our error bound, so we can just calculate our error abound. So all we have to do now is substitute our value of 𝑀 into our error bound. We get the absolute value of our remainder term 𝑅 two is less than or equal to the absolute value of five divided by 3456 multiplied by 8.5 minus eight all raised to the power of two plus one divided by two plus one factorial. And if we were to calculate this expression, we will get it’s equal to five divided by 165888.

But remember, the question is not asking us to give this as an exact value; we need to give this in scientific form to three significant figures. So we’ll start by writing out the decimal expansion of this. It’s equal to 0.00003014 and this continues. And we want to write this to three significant figures. First, remember, the initial band of zeros don’t count. We want the first three digits after this initial band of zeros. In this case, it’s three, zero, and one.

Next, we need to look at our first omitted digit to see if we need to round up. Since our first omitted digit is four, which is less than five, we don’t need to round up. Then, we’ll just write this in scientific notation to three significant figures. It’s approximately equal to 3.01 times 10 to the power of negative five.

And this gives us our final answer. The absolute value of 𝑅 two will be less than or equal to 3.01 times 10 to the power of negative five. Therefore, we were able to find the error bound when using the second Taylor polynomial for the cube root of π‘₯ centered at π‘₯ is equal to eight to approximate the value of 𝑓 evaluated at 8.5. In scientific form to three significant figures, we got 3.01 times 10 to the power of negative five.

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