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Question Video: Finding the Family of Functions That Have a Given Third Derivative Mathematics • 12th Grade

Determine the family of functions 𝑓 for which 𝑓‴(𝑑) = 3 sin 𝑑 βˆ’ 2.


Video Transcript

Determine the family of functions 𝑓 for which 𝑓 triple prime of 𝑑 equals three sin 𝑑 minus two.

What does the notation 𝑓 triple prime of 𝑑 mean? It’s the notation for the third derivative. It means the function 𝑓 of 𝑑 has been differentiated three times to get 𝑓 triple prime of 𝑑. So we need to do the reverse of differentiation or integration three times to get 𝑓 of 𝑑. So let’s get started and find the first antiderivative of three sin 𝑑 minus two. In the same way that we differentiate term by term, we can integrate term by term. So we’re integrating three sin 𝑑 and two. It might be helpful to recall some useful trick derivatives. Sin 𝑑 differentiates to cos 𝑑. Cos 𝑑 differentiates to negative sin 𝑑. Negative sin 𝑑 differentiates to negative cos 𝑑. And negative cos 𝑑 differentiates to sin 𝑑. This is helpful because, for integration, the arrows go exactly the opposite way.

Let’s also remind ourselves of integration of other functions. If we have a function such as four π‘₯ squared plus three π‘₯ plus five, we can differentiate this term by term. We start with four π‘₯ squared and multiply the coefficient four by the power two to get the new coefficient of eight. And then, we reduce the power by one to get eight π‘₯. Three π‘₯ is three π‘₯ to the power of one. So this differentiates to three. It’s going to be helpful to us at this point to note that three π‘₯ differentiates to three. And so three integrates to three π‘₯. Five is a constant. So it differentiates to zero.

So that’s differentiation. But what about antidifferentiation or integration? Well, it’s the reverse of differentiation. Instead of multiplying the coefficient by the power and subtracting one from the power, we increase the power by one and divide the coefficient by that new power. Because constants differentiate to zero, we account for that by adding a constant to our antiderivative. We often represent the constant with 𝐢. But really we can use any letter.

So going back to our question, we want the antiderivative of three sin 𝑑 minus two. And this will give us 𝑓 double prime of 𝑑. If we check our diagram, we see that sin 𝑑 integrates to negative cos 𝑑. So three sin 𝑑 integrates to negative three cos 𝑑. And two integrates to two 𝑑 as we saw in our example. And we must include the constant of integration, 𝐢. And now we integrate again to find 𝑓 prime of 𝑑. Our diagram shows that negative cos 𝑑 integrates to negative sin 𝑑. So negative three cos 𝑑 integrates to negative three sin 𝑑. To integrate two 𝑑, we follow our rules for antidifferentiation. Increasing the power of 𝑑 by one gives us the new power of two. And then, we divide the coefficient two by the new power of two, which is just one. So this is just 𝑑 squared. We’ve seen how to integrate a constant. So to integrate 𝐢, we just get 𝐢𝑑. This is in the same way that two integrated to two 𝑑, 𝐢 integrates to 𝐢𝑑.

And finally, because we’ve integrated, we have to include a constant of integration. So this time let’s call it 𝐷. So we’ve got 𝑓 prime of 𝑑. And we have to keep going because, remember, we’re trying to get 𝑓 of 𝑑. From our diagram, negative sin 𝑑 integrates to cos 𝑑. So negative three sin 𝑑 integrates to three cos 𝑑. To integrate 𝑑 squared, we increase the power by one to get the new power of three. And then, we divide by the new power. So we get 𝑑 cubed over three.

To integrate 𝐢𝑑, we follow our rules for antidifferentiation again. We increase the power by one and divide by the new power to give us 𝐢𝑑 squared over two. However, 𝐢 is an unknown constant. So dividing it by two means it’s still an unknown constant. So we can just write 𝐢𝑑 squared. 𝐷 is a constant. So that integrates to 𝐷𝑑. And our final step is to add a constant of integration, which we can call 𝐸 this time. So now we have 𝑓 of 𝑑.

This is a family of functions because there’s lots of possibilities for the constants 𝐢, 𝐷, and 𝐸. All the possibilities for 𝐢, 𝐷, and 𝐸 create this family of functions. For a final answer, we often write the term with the highest power at the beginning of the function. In this case, it’s 𝑑 cubed over three.

So our final answer is negative 𝑑 cubed over three plus three cos 𝑑 plus 𝐢𝑑 squared plus 𝐷𝑑 plus 𝐸.

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