Original subtitles

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Today we're going to talk about how to find objects.

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Given the function second derivative and a couple of initial conditions we complete this problem will

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think the integral of second derivative to find the first derivative then take the integral of the first

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derivative find the original function and then plug in our initial conditions.

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In this particular problem we've been asked to find F X if double prime of x is equal to 12 x squared

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plus 6 X minus 4.

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And given the initial conditions that EF 0 0 0 is equal to 4 and 1 is equal to 1.

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So our first step is to start with F double prime IVAX and start picking antiderivatives right.

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We know that f of x is our essentially original function f prime is its first derivative and of double

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prime is its second derivative which means therefore starting with F double prime of X will need to

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take the integral of double prime of X to get two f prime of X the first derivative and then take the

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integral again to get from prime of X back to f of x.

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So let's start with the second derivative f double prime of X and say that that is equal to 12 x squared

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plus 6 X minus 4.

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Now to find f prime of X the first derivative we need to take the integral of f a prime of x which we

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know to be 12 x squared plus 6 x minus 4.

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And as with any integral we need that DX notation there.

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So is this a simple polynomial integral.

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Remember that when we're taking the integral or the antiderivative of this function here we're just

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going to use a power rule.

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So you take the integral we have x squared.

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Here we're going to be adding one to the exponent.

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So two plus one in the exponent gives us three and then we divide the coefficient 12 by the new exponent

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3.

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Same thing here with the 6 x ray.

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So we have basically X to the first power.

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Right.

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X to the first power.

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So we'll add 1 to the exponent one plus one gives us two.

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So we'll get X squared and then we'll divide the coefficient 6 by the new exponent which is 2.

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Finally we get to form and basically here we have four times x to the zero power.

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Right.

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X to the zero or anything to the zero for that matter is equal to 1 4 times 1 is just for this term

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as essentially you can think of it as 4 x to the 0.

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So it becomes the same then as the last two terms that we've done.

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So we have X to the zero.

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Again we'll add 1 to the exponent zero plus 1 gives us 1.

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So there's our exponent and then we'll take the coefficient for and divide it by the new exponent which

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is 1.

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And as with any integral whenever you take the integral you have to go ahead and add what we call C

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which is the constant of integration to account for a constant that may have been in f prime of X that

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got lost when we took the derivative to get at a prime of x.

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So now that we have a prime of x we just need to simplify 12 divided by 3.

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Here is 4.

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So we get 4 x cubed 6 over 2 will give us 3 x squared.

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And obviously this here just becomes for X and then we keep our C for the constant integration so that

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f prime of x.

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Now we need to go ahead and get to the original function f of x.

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So to get f of x we need to take the integral again of the first derivative function as prime of x so

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we'll be taking the integral of 4 x cubed plus 3 x squared minus 4 x plus C and again at Dx notation.

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So we'll do the same thing we did last time for 4 x cubed here to take the integral again.

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Add 1 to the exponent three plus one gives us 4 and we'll divide our coefficient for our new exponent

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4.

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Same thing here.

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You've got x squared well add 1 to the exponent to get execute and then we'll divide the coefficient

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by the new exponent for negative for X here we have X to the first power.

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So add 1 and we'll get X squared and then we'll divide the coefficient or the new exponent.

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See here is just a constant.

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This basically takes over the same form that the forehead last time.

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This is c times x to the 0 or c times 1 because anything raised to the zero power is just one.

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So we add one to the exponent and the x to the first the coefficient is C C is just a constant.

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Like any number here just like for was up here in our original second derivative.

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So we get C divided by 1.

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And because we took integral again what we have to add another constant to distinguish it from the constant

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we added before we'll call this one D instead of C.

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So now we need to simplify as much as we can for forgiveness 1.

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So we'll just get X to the fourth again three over three will just be one.

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So we'll get x cubed and then negative four halves gives us a negative 2 x squared plus C X plus the

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so this here is our original function f of x.

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Now we need to do is use our initial conditions of zero equals four and a half of one equals one to

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find C and D are constants.

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So the way that we'll do that because these are both the original function f of zero as opposed to prime

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of zero or double prime of zero.

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We're going to be plugging them into our original function here f of x.

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So what this tells us to do is plug in zero for X and set it equal to 4.

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So we'll say and we'll be doing it here with this equation that we simplified for objects.

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So we'll set it equal to 4 and plug in zero.

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So we'll get 0 to the fourth power plus 0 cubed minus two times zero squared plus C times zero plus

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the.

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And that'll be our first equation.

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We can simplify that obviously because we'll get all of these to cancel since they'll all be zero.

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And what we're left with here is four equals D.

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Now we need to use our second initial condition to solve for C.

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So we'll plug in one for X and that the whole thing equal to one.

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So you get one equals one to the fourth power plus one to the third power minus two times 1 squared

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plus C times 1 plus the.

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And what we'll get when we simplify this is one equals.

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This will just be one plus one minus two plus C plus the.

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So simplify again.

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One plus one minus two is two minus two.

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So these were all cancell with one another and we're just left with C plus D.

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We already solved for D.

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Remember that we got the equals four.

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We can go ahead and make that substitution for D.

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And then to solve for C We'll just subtract C from both sides and we'll see that negative 3 is equal

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to see.

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So now we have solutions for C and D.

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And what we can do is plug them back into our original function to get our final answer.

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So our final answer for f objects will be f objects is equal to x to the fourth plus x cubed minus 2

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x squared minus 3 because we get negative 3 for C minus 3 x plus 4 for B so plus 4.

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And that's it.

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That's our original function f of x given the second derivative f double prime of X and these two initial

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conditions.