- [Voiceover] Find the value of k
for which f has a critical point at x equals negative five.
All right, so let's just remind ourselves
what f of x and f prime of x were.
They gave it to us at the top,
f of x is equal to one over x squared minus kx,
and then f prime of x is equal to all of this business.
Let me rewrite it,
let me rewrite it down here,
so f,
f of x is equal to one over
x squared minus kx
and f prime of x is equal to
k minus two x
over x squared minus kx,
whole quantity squared.
And we wanna find the value of k
for which f has a critical point at x equals negative five,
so f has a critical point at x equals negative five.
This means that x,
that means that x equals negative five is in domain,
so this is in domain,
which means our function itself
is defined at x equals negative five,
and it means that f prime of negative five
is equal to zero or undefined,
or undefined.
A critical point is a member of the functions domain
where the derivative is equal to zero,
or it's undefined.
So let's evaluate f prime of negative five.
F prime of negative five in terms of k
is going to be equal to k minus two times negative five,
minus two times negative five,
all of that over
negative five squared minus k
times negative five,
and then we want to square all of this,
and so this is going to be equal to k plus 10,
k plus 10 over,
we have 25, what is this?
25 plus five k,
plus five k squared.
So what value of k makes f prime
of negative five equal to zero?
Well, f prime of negative five
is equal to zero
if k is equal to negative ten.
So that's a value of k for which f has
a critical point at negative five.
Now, you might be saying,
well, what values of k?
That's the value of k that makes
the function equal zero,
it makes the numerator equal zero,
and therefore makes the whole function equal zero.
But why can't I pick a value of k that makes the derivative,
or it makes the numerator of the derivative zero,
and so therefore makes the derivative equal zero,
I think I said function,
not the derivative of the function.
But you might be saying,
well, why can't I pick a value of k
that makes the derivative undefined?
And so, you can think of what that is,
what would make undefined,
well, if 25 plus five k is equal to zero,
well, then this is going to be,
you're gonna have zero squared,
and divide it by zero,
it's going to be undefined,
so you could say f prime of negative five
undefined,
undefined if k is equal to what?
Negative five, right?
Five times negative five plus 25 is zero,
but this,
if k is equal to negative five,
then this can't be a critical point anymore,
x equals negative five can't be a critical point
because it won't be in it's domain anymore.
If k is equal to negative five,
then f of x would be equal to one
over x squared plus five x,
plus five x,
and then f will not be defined at negative five,
and so the negative five couldn't be a critical point
because it's not even in the domain.
So the important thing is,
in order to be a critical point,
it has to be in the domain,
and the derivative at that point
has to be equal to zero undefined.
I can get the derivative to be undefined at negative five,
if we said k equals negative five,
but with we said k equals negative five,
then x equals negative five will no longer be in the domain,
so we wanna go with the k that just makes the numerator
of our derivative equal to zero,
or sets our entire derivative equal to zero.