Video: Finding the Unknown Coefficients in the Expression of a Function given the Value of the Second and Third Derivatives of a Function

Given that 𝑦 = π‘Žπ‘₯Β³ + 𝑏π‘₯Β², 𝑦‴ = βˆ’18, and [𝑑²𝑦/𝑑π‘₯Β²]_(π‘₯ = 2) = βˆ’14, find π‘Ž and 𝑏.

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Video Transcript

Given that 𝑦 equals π‘Žπ‘₯ cubed plus 𝑏π‘₯ squared, the third-order derivative of 𝑦 is equal to negative 18, and 𝑑 two 𝑦 over 𝑑π‘₯ squared when π‘₯ is equal to two is equal to negative 14, find π‘Ž and 𝑏.

So the first thing we need to do in this question is actually find our first-, second-, and third-order derivatives. So we’re gonna start with our first-order derivative. And just to remind us how we actually differentiate if we have a function in π‘₯, then if it’s in the form π‘Žπ‘₯ to the power of 𝑏, then our first-order derivative is gonna be equal to π‘Žπ‘π‘₯ to the power of 𝑏 minus one. So you multiply the coefficient by the exponent and reduce the exponent by one.

Okay, so let’s use that and let’s find the first-order derivative of our function. So the first term is gonna be three π‘Žπ‘₯ squared. And that’s because we multiplied our exponent three by a coefficient π‘Ž. So we get three π‘Ž and then we’ve reduced the exponent by one. So we get π‘₯ squared. So then, the second term is just plus two 𝑏π‘₯. Again, we’ve differentiated that term. Okay, so that’s our first-order derivative.

What we’re gonna move onto now is going to differentiate again to find our second-order derivative. So our second-order derivative which I’ve shown here with 𝑦 prime prime. And all as we’ve got it in the question 𝑑 two 𝑦 over 𝑑π‘₯ squared is gonna be six π‘Žπ‘₯ plus two 𝑏. So again, we just differentiate that in the normal way. So we have three π‘Ž multiplied by two β€” our coefficient multiplied by our exponent β€” and then π‘₯ to the power of two minus one, which is just π‘₯ β€” so π‘₯ to the power of one. Okay, great, so now we’ve found our second-order derivative.

So now, great, we can actually move on and find our third-order derivative. And all we do to find this is actually differentiate again, which is just gonna leave us with six π‘Ž because if we differentiate six π‘Žπ‘₯, we’re just left with six π‘Ž. And if we differentiate plus two 𝑏, then it’s just going to be zero because if we differentiate any number that hasn’t got an π‘₯ term in, it just goes to zero. Okay, so now we’ve done that β€” we’ve got our first-, second-, and third-order derivatives β€” let’s see what we do next.

Well, let’s start with this bit of information. We know that our third-order derivative is gonna be equal to negative 18. So we can now set up an equation. And we can actually set up that new equation because we can say that our third-order derivative is equal to negative 18. So can sub that in. So we get negative 18 is equal to six π‘Ž. Okay, so therefore, we can say that π‘Ž is going to be equal to negative three. And we’ve got that because we’ve actually divided through by six. And also, we just flipped it just so we got the π‘Ž on the left-hand side.

So fantastic, we’ve now found π‘Ž. So now, what we need to do is move on and have a go at finding 𝑏. And in order to do that, what we’re gonna use is this bit of information here. Because if we take a look at this bit of information, we’ve got that 𝑑 two 𝑦 over 𝑑π‘₯ squared β€” so our second-order derivative β€” is equal to negative 14 when π‘₯ is equal to two. So therefore, in order to actually find 𝑏, what we can do is substitute in π‘₯ is equal to two and 𝑦 prime prime is equal to negative 14. And this is gonna give us that negative 14 is equal to six π‘Ž multiplied by two plus two 𝑏.

But we’ve now got something else. We’re gonna substitute in because we know π‘Ž. So we can substitute in π‘Ž is equal to negative three. So therefore, we get that negative 14 equals six multiplied by negative three multiplied by two plus two 𝑏. So we have negative 14 equals negative 36 plus two 𝑏. So then, what we do is we actually add 36 to each side. So we have 22 equals two 𝑏. So therefore, if we divide by two, we get 𝑏 is equal to 11.

So that means that we can say that given that 𝑦 equals π‘Žπ‘₯ cubed plus 𝑏π‘₯ squared which third-order derivative of 𝑦 is equal to negative 18 and 𝑑 two 𝑦 over 𝑑π‘₯ squared when π‘₯ is equal to two is equal to negative 14, π‘Ž is equal to negative three and 𝑏 is equal to 11.

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