104–107. Functions from derivatives Find the function f with t...

Question

104–107. Functions from derivatives Find the function f with the following properties.

h'(x) = (x⁴ -2) /(1 + x²) ; h (1) = -(2/3)

Accepted Answer

Hi, everyone, let's take a look at this practice problem. This problem says to find the function G of X, such that G of X is equal to the quantity of X of 4th plus 3, and quantity divided by the quantity of 1 + X squared in quantity, and G of 0 is equal to 2. Now, in order to find a function G of X, we just need to integrate G of X with respect to X. So that means that G of X. It's going to be equal to the integral of the quantity of X 4th, plus 3 in quantity, divided by the quantity of 1 plus X squad in quantity DX. And in order to. Get this integrated into a form that we can actually integrate, we actually need to do long division for these polynomials. So, we're gonna take 1 + X2d and divide it into. The quantity of x 4 + 3. And so during the long division, we'll see that X2 will go into X the 4th, X2 times. So, we'll have X2 multiplied by the quantity of 1 + X2, and that will give us X4 plus X2, and we'll subtract that quantity from X 4 + 3. In doing so, we will get minus X2 + 3. And X2 will go into minus X2 minus 1 times, so we'll subtract 1 in the quotient. And so we'll have -1 multiplied by the quantity of 1 minus X2, and this will give us minus X2 minus 1. And we're gonna subtract this quantity from minus X2 + 3. In doing so, we will actually have a remainder of 4. That means that we can rewrite our integral as being equal to the integral of the quantity of X2 minus 1. Plus 4 divided by the quantity of 1 plus x 2 in quantity, and that is all multiplied by DX. So now, we can perform the integration by integrating each of these terms individually. So doing so, this is going to be equal to. For the first term, we need to integrate X2, and recall that when we integrate X2 with respect to X, we get X cubed divided by 3. Then we'll have minus for the 2nd term, we're integrating 1 with respect to X, which will give us X. And then we'll have plus, we'll go ahead and factor out the 4, and what we could be left with is 1 divided by the quantity of 1 plus X2, and when I integrate 1 divided by the quantity of 1 plus X2. With respect to X, I get the inverse tangent. Of X. And then, each one of these integrations will give us a constant, and so we'll just have a total integration constant of plus C at the end. Next, we need to actually find the value of this integration constant, and we're told in the problem that when X is equal to 0, G of X is equal to 2. So we'll use that information to find the integration constant. So for G of X, we will have 2, and this is going to be equal to 2. And here on our expression for G of X, we'll substitute N X equal to 0. So we'll have 0 cubed divided by 3 minus 0 + 4 multiplied by the inverse tangent. Of 0. Plus C. Now, all the terms on the right hand side are 0 except for C. So that means that C is going to be equal to 2. So, our final expression for G of X is going to be equal to 1 divided by 3, multiplied by X cubed minus X plus 4 multiplied by the inverse tangent of X. Plus 2. So this here is the answer that we were looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

104–107. Functions from derivatives Find the function f with the following properties.

h'(x) = (x⁴ -2) /(1 + x²) ; h (1) = -(2/3)

Key Concepts

Derivatives

Integration

Initial Conditions

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