hi everyone today we're going to talk about how to find f of x given the functions second
derivative and a couple of initial conditions to complete this problem we'll take the integral
of the second derivative to find the first derivative then take the integral of the first
derivative to find the original function and then plug in our initial conditions let's
take a look in this particular problem we've been asked to find f of x if f double prime
of x is equal to twelve x squared plus six minus four and given the initial conditions
that f of zero is equal to four and f of one is equal to one so our first step is to start
with f double prime of x and start taking antiderivatives right we know that f of x
is our essential original function f prime is its first derivative and f double prime
is its second derivative which means that if we're starting with f double prime of x
we'll need to take the integral of f double prime of x to get to f prime of x the first
derivative and then take the integral again to get from f prime of x back to f of x so
let's start with the second derivative f double prime of x and say that that is equal to twelve
x squared plus six x minus four now to find f prime of x the first derivative we need
to take the integral of f double prime of x which we know to be twelve x squared plus
six x minus four and as with any integral we need that dx notation there so this is
a simple polynomial integral remember that when we're taking the integral or the antiderivative
of this function here we're just going to use power rule so to take the integral we
have x squared here we're going to be adding one to the exponent so two plus one in the
exponent gives us three and then we divide the coefficient twelve by the new exponent
three same thing here with the six x right so we have basically x to the first power
right x to the first power so we'll add one to the exponent one plus one gives us two
so we'll get x squared and then we'll divide the coefficient six by the new exponent which
is two finally we get to four and basically here we have four times x to the zero power
right x to the zero or anything to the zero for that matter is equal to one four times
one is just four this term is essentially you can think of it as four x to the zero
so it becomes the same then as the last two terms that we've done so we have x to the
zero again we'll add one to the exponent zero plus one gives us one so there's our exponent
and then we'll take the coefficient four and divide it by the new exponent which is one
and as with any integral whenever you take the integral you have to go ahead and add
what we'll call c which is the constant of integration to account for a constant that
may have been in f prime of x that got lost when we took the derivative to get f double
prime of x so now that we have f prime of x we just need to simplify twelve divided
by three here is four so we get four x cubed six over two will give us three x squared
and obviously this here becomes four x and then we keep our c for the constant of integration
so that's f prime of x now we need to go ahead and get to the original function f of x so
to get f of x we'll need to take the integral again of the first derivative function f prime
of x so we'll be taking the integral of four x cubed plus three x squared minus four x
plus c and again at dx notation so we'll do the same thing we did last time for four x
cubed here to take the integral again we'll add one to the exponent three plus one gives
us four and we'll divide our coefficient four by our new exponent four same thing here we've
got x squared we'll add one to the exponent to get x cubed and then we'll divide the coefficient
by the new exponent for negative four x here we have x to the first power so we'll add
one and we'll get x squared and then we'll divide the coefficient by the new exponent
c here is just a constant this basically takes over the same form that the four had last
time this is c times x to the zero or c times one because anything raised to the zero power
is just one so we add one to the exponent and we get x to the first the coefficient
is c c is just a constant like any number here just like four was up here in our original
second derivative so we get c divided by one and because we took an integral again we have
to add another constant but to distinguish it from the constant we added before we'll
call this one d instead of c so now we need to simplify as much as we can four over four
gives us one so we'll just get x to the fourth again three over three will just be one so
we'll get x cubed and then negative four halves gives us a negative two x squared plus c x
plus d so this here is our original function f of x now all we need to do is use our initial
conditions f of zero equals four and f of one equals one to find c and d our constants
so the way that we'll do that because these are both the original function f of zero as
opposed to f prime of zero or f double prime of zero we're going to be plugging them into
our original function here f of x so what this tells us to do is plug in zero for x
and set it equal to four so we'll say and we'll be doing it here with this equation
that we simplified for f of x so we'll set it equal to four and plug in zero so we'll
get zero to the fourth power plus zero cubed minus two times zero squared plus c times
zero plus d and that will be our first equation we can simplify that obviously because we'll
get all of these to cancel since they'll all be zero and what we're left with here is four
equals d now we need to use our second initial condition to solve for c so we'll plug in
one for x and set the whole thing equal to one so we'll get one equals one to the fourth
power plus one to the third power minus two times one squared plus c times one plus d
and what we'll get when we simplify this one equals this will just be one plus one minus
two plus c plus d so simplify again one plus one minus two is two minus two so these will
all cancel with one another and we're just left with c plus d we've already solved for
d remember that we got d equals four so we can go ahead and make that substitution for
d and then just solve for c we'll just subtract c from both sides and we'll see that negative
three is equal to c so now we have solutions for c and d and what we can do is plug them
back into our original function to get our final answer so our final answer for f of
x will be f of x is equal to x to the fourth plus x cubed minus two x squared minus three
because we got negative three for c minus three x plus four for d so plus four and that's
it that's our original function f of x given the second derivative f double prime of x
and these two initial conditions so I hope you found that video helpful if you did like
this video down below and subscribe to be notified of future videos