2015 AP Calculus BC 5c

Find the value of K for which F has a critical point at X = -5.

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 ( \frac{1}{x^2 - Kx} ) and then F prime of X is equal to all of this business.

Let me rewrite it down here. So F of X is equal to ( \frac{1}{x^2 - Kx} ) and F prime of X is equal to ( \frac{K - 2x}{(x^2 - Kx)^2} ). We want to find the value of K for which F has a critical point at X = -5.

This means that X = -5 is in the domain. This means our function itself is defined at X = -5, and it means that F prime of -5 is equal to zero or undefined. A critical point is a member of the function's domain where the derivative is equal to zero or it's undefined.

So let's evaluate F prime of -5 in terms of K.

F prime of -5 in terms of K is going to be equal to ( \frac{K - 2 \cdot -5}{(-5)^2 - K \cdot -5} ), and then we want to square all of this.

So this is going to be equal to ( K + 10 ) over ( 25 + 5K ).

So what value of K makes F prime of -5 equal to zero? Well, F prime of -5 is equal to zero if K is equal to -10.

So that's the value of K for which F has a critical point at -5. 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 to 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 therefore makes the derivative equal to 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?” So you could think of what that is.

What would make this undefined? Well, if ( 25 + 5K = 0 ), then this is going to be undefined. You're going to have ( 0 ) squared divided by ( 0 ), it's going to be undefined.

So you could say F prime of 5 is undefined if K is equal to -5, right? ( 5*5 + 5 = 0 ).

But if K is equal to -5, then this can't be a critical point anymore at X = 5. It can't be a critical point because it won't be in its domain anymore.

If K is equal to -5, then F of X would be equal to ( \frac{1}{x^2 + 5x + 5} ), and then F will not be defined at 5.

So, neg5 couldn't be a critical point because it's not even in the domain.

The important thing is, in order to be a critical point, it has to be in the domain, and the derivative of that point has to be equal to zero or undefined. I can get the derivative to be undefined at 5 if we set K to -5, but if we set K to -5, then X = 5 will no longer be in the domain.

So we want to go with the K that just makes the numerator of our derivative equal to zero or sets our entire derivative equal to zero.