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# 2015 AP Calculus BC 5c

## Video transcript

find the value of K for which F has a critical point at x equals negative 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 1 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 1 over x squared minus KX and F prime of X is equal to K minus 2 x over over x squared minus KX whole quantity squared and we want to find the value of K for which F has a critical point at X is equal to negative 5 so f has a critical point at x equals negative 5 this means this means that X that means that x equals negative 5 is in domain so this is in domain which means our function itself is defined at x equals negative 5 and it means that f prime of negative 5 is equal to 0 or undefined or undefined a critical point is a member of the functions domain where the derivative is equal to 0 or it's undefined so let's evaluate F prime of negative 5 f prime of negative 5 in terms of K is going to be equal to K minus 2 times negative 5 minus 2 times negative 5 all of that over negative 5 squared minus K times negative 5 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 5 K 25 plus 5 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 0 if K is equal to negative 10 so that's a value of K for which F has a critical point at negative 5 now you might be saying well what values of K that that's the value of K that makes the function equal zero it makes a numerator equals zero and therefore makes the whole function equal to zero but why can't I pick up K a value of K that makes that makes the derivative or it makes the it makes the numerator of the derivative zero and so therefore it makes the derivative equal to zero I think I said function not but 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 could think of what that is what would make this undefined well if 25 + 5 k is equal to zero well then this is going to be you're gonna have zero squared u divided by zero Square to be undefined so you could say F prime of negative five undefined undefined if K is equal to what negative 5 right 5 times negative 5 plus 25 is zero but this if if K is equal to negative five then this can't be a critical point anymore x equals negative 5 can't be a critical point because it won't be in its domain anymore if K is equal to negative five then f of X would be equal to 1 over x squared + 5 X + 5 X and then F will not be defined at negative 5 and so negative 5 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 to that point has to be equal to zero undefined I can get the derivative to be undefined at negative 5 if we set K equals negative 5 but if we set K equals negative 5 then x equals negative 5 will no longer be in the domain so we what we want to go with the K that just makes the numerator of our derivative equal to 0 or sets our entire derivative equal to 0
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