hi everyone today we're going to talk about how to find f of x given the functions third
derivative to complete this problem we'll work backwards taking the integral of the
third derivative to get the second derivative the integral of the second derivative to get
the first derivative and then the integral of the first derivative to find the original
function let's take a look in this particular problem we've been asked to find f of x if
x triple prime or the third derivative of x is equal to x minus the square root of x
so in order to find our way from f triple prime or the third derivative of x back to
f of x we're going to have to work our way backwards so we'll start with the third derivative
of x and we'll try to find the second derivative of x once we've found it we'll find the first
derivative and then use the first derivative to find the original function so we'll start
with the third derivative of x and in order to find the second derivative of x f double
prime we'll need take the integral or the anti-derivative of the third derivative so
x minus the square root of x and whenever you have integral notation like this you're
going to need a dx corresponding notation to go with it so let's go ahead and make a
change in this integral to make it easier on ourselves we're going to be taking the
integral of x I want to convert the square root of x here to x to the one half which
is the same thing because now we can use power rule to easily take the integral of this term
so to take the integral remember that we're going to use simple power rule here we have
x to the first power x to the one to take the integral we'll add to the exponent we'll
get one plus one equals two and we'll divide the coefficient which right now is just one
divided by new exponent which already found to be two so that's how we use power rule
to take the integral of simple polynomial terms like this so now we need to take the
integral of negative x to the one half in order to do that we'll add one to the exponent
so one half plus one will give us x to the three halves and then we'll divide the coefficient
which is currently one by the new exponent so divided by three halves when we simplify
this we'll get one half x squared minus two thirds x to the three halves but then keep
in mind that whenever we take an integral we always have to add c or the constant of
integration so we need to add c to account for the constant of integration here and of
course that would carry down into our simplification to what we solved for here is the second derivative
of f of x now we need to go ahead and find the first derivative so the first derivative
will f prime of x and that will be equal to the integral of the second derivative so one
half x squared minus two thirds x to the three halves plus c dx so again we'll use power
rule to take the integral of this term by term we have x squared here we'll add one
to exponent two plus one gives us three and then we divide the coefficient one half by
three one half divided by three is one sixth so we get one sixth there now we have negative
two thirds x to the three halves take the integral we'll add one to exponent three halves
plus one will give us five halves so x to five halves and then we'll divide negative
two thirds by five halves the new exponent but negative two thirds divided by five halves
is the same thing as negative two thirds times two fifths right remember instead of dividing
by a fraction we can multiply by its inverse so this is negative two thirds times two fifths
and then when we take the integral of c we'll just get c x because remember that c is just
the constant and this is basically c times x to the zero power x to the zero and anything
raised to the zero is one so essentially we haven't changed this at all because we're
just saying c times one which is c this now follows though the same rules as the other
two terms so since we have x to the zero we'll add one to exponent and get x to the first
power and then divide the coefficient c by the new exponent one but of course because
we've integrated we also want to add another constant since we've already used c to account
for a constant we add d instead now we just need to simplify this as much as we can so
we'll get one sixth x to the third we have negative two thirds times two fifths so we'll
get minus four over fifteen x to five halves plus cx plus d and that is our first derivative
now in order to finally get back to our original function f of x we will take the integral
of our answer there so f of x will the integral of one sixth x to the third minus four over
fifteen x to the five halves plus cx plus d and as always will have our dx here to take
the integral we'll do the same thing we've been doing we'll add one to the exponent so
x to the third becomes x to the fourth and now we have one sixth divided twenty four
which will be one twenty fourth taking the integral of this x to the five halves well
add one to the exponent five halves plus one gives us seven halves so x to the seven halves
we'll get negative four over fifteen instead of dividing by seven halves we'll multiply
by the inverse two sevenths then we get plus we add one to the exponent this is essentially
x to the first so we'll get x squared and we'll divide the coefficient c by the new
exponent which is two and then same here d this is essentially d times x to the zero
we add one to the exponent and we get x to the first power we divide d by the new exponent
so we get d over one but because we're integrating of course we have to add another constant
of integration this time we'll call it e so when we simplify this we'll get our final
answer for f of x which is just going to be one over twenty four x to the fourth negative
four fifteenths time two sevenths will be negative eight over one over one o five x
to the sevenths halves plus c over two x squared plus dx plus e and that's it that's our final
answer so I hope you found that video helpful if you did like this video down below and
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