Original subtitles

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It we're going to talk about how to find f of x even the function second derivative to complete this

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problem we'll work backwards and take the integral of the second derivative to find the first derivative

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and then the integral of the first Serota to find the original function.

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In this particular problem we've been asked to find f of x f f l prime.

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The second derivative of x is equal to two thirds X to the two thirds.

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So in order to find f of x the original function from prime of X or the second derivative we need to

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work backwards.

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What we'll do is take the integral of prime of X to find the first tentative prime of X and then take

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the integral of that to find other backs.

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If we have the second derivative f double prime of X equal to two thirds X to the two thirds and to

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find the first derivative of prime of X will take the integral of the second derivative.

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Two thirds X to the two thirds.

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And when we take the integral we have to have the corresponding DX notation here.

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So you take the integral we'll just use simple power rule and because we have X to the two thirds here

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we'll add 1 to the exponent.

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So two thirds plus one or we can say two thirds plus three thirds will give us 5 thirds so we'll get

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X to the 5 thirds and then we want to take the coefficient two thirds and divide it by the new exponent

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5 thirds.

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So two thirds divided by five thirds.

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But instead of picking a fraction divided by a fraction we can say two thirds and instead of dividing

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by five thirds that's the same thing as multiplying by the inverse of bibe thirds which is three fifths.

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So that is the integral of the second derivative.

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But of course whenever we take an integral we have to add c so we get plus C here and all we need to

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do is simplify.

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Notice obviously that the three in the numerator and denominator here will cancel.

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So we'll get to Firth's X to the five thirds plus C and that is f prime of x are first derivative function.

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Now we need to take the integral of that to get back to at the max.

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So ever x will be equal to the integral of our first derivative function.

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So to fix X to the 5 third's plus C and of course as always our dx notation.

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Now in order to take this in a roll do the same thing we did last time.

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Add 1 to the exponent so x to the 530.

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Here we add 1 to 5 thirds 5 thirds plus one will just give us 8 thirds.

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So we'll get X to the 8 thirds.

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And now we want to take two fifths the coefficient and divide it by the new exponent.

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So two fifths divided by eight thirds is the same thing as two fifths times three eighths So multiply

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them.

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And now see we can treat as a constant.

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This is basically the same thing as saying See times x to the zero power X to zero or anything to the

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zero power for that matter is equal to 1.

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So this is the same as saying See times 1.

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Or just see.

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So by multiplying by X to zero we haven't changed anything in it.

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Now allows to treat this the same way that we did this X of the five third's here.

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We'll just add one to the exponent so we'll get zero plus one just gives us one and then we'll divide

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C by the new exponent so see you divided by 1 is what we'll get.

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And because we took another integral we need to account for the constant of integration and add C again.

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We want to distinguish it because we've already added C..

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So instead of using C we'll use the next variable and call it D.

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So now we just need to simplify this as much as we can to get our final answer.

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Well get f of x is equal to two fifths times 8 will give us 6 over 40 which is the same thing as 3 over

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20 x to the 8 thirds plus see X plus the.

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And that's it.

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That is our original function f of x.

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If the second derivative f double prime of x is equal to two thirds X to the two thirds.