Find the critical points of f. Assume a is a constant. f(x)...

Find the critical points of f. Assume a is a constant. f(x) = x?x - a Select the correct choice below and, if necessary, fill in the answer box within your choice. A. For all values of a, the critical point(s) is(are) x = [ ]. (Use a comma to separate answers as needed.) B. If a ? 0, the critical point(s) is(are) x = [ ]. If a > 0, there is no critical point. (Use a comma to separate answers as needed.) C. If a < 0, the critical point(s) is(are) x = [ ]. If a ? 0, there is no critical point. (Use a comma to separate answers as needed.) D. If a ? 0, there is no critical point. If a > 0, the critical point(s) is(are) x = [ ]. (Use a comma to separate answers as needed.) E. If a < 0, there is no critical point. If a ? 0, the critical point(s) is(are) x = [ ]. (Use a comma to separate answers as needed.) F. There are no critical points.

Transcript

00:01 We're going to find the critical points of f, for that we assume that a is a constant, and the function f is x times square root of x minus a.

00:14 So, first of all, we know that the domain of f is the real numbers for which x is created than or equal to a, for the square root of x minus a to be defined on the real numbers.

00:37 Numbers.

00:38 So that's the domain and there is where we can find the critical points.

00:46 So let's calculate the derivative of f respect to x and that is we have a product x times x squared root of x minus a which can be written as x minus a to the one half.

01:01 And written this way we're going to find the derivative of a product so we get derivative of x respect to x times the square root of x minus a plus x times derivative respect to x of x minus a to the one half.

01:27 Good.

01:27 So that is equal to this derivative is equal to 1.

01:31 So we get square root of x minus a plus x times.

01:37 And this derivative here is 1 .5 times x minus a to the 1 half minus 1 times derivative respect to x of x minus a applying the chain rule.

01:52 And so we get square root of x minus a plus x times 1⁄2 times x minus a.

02:05 1⁄2 minus 1 .5 is negative 1⁄2 times and this derivative here is equal to 1 because the derivative of x respect to x is 1 and the derivative of a is 0 because a is a constant.

02:18 So we get times 1 and so we get square root of x minus a plus then we have x over 2 square root of x minus a because x minus a to the negative 1 half is the same as 1 over x minus a to the one half that is one over square root of x minus a that is square root of x minus a gets passed to the denominator and we have x in the numerator and with these two here we have in the denominator also so we have this and so we can say that the derivative of f respect to x is this formula here square root of x minus a plus x divided by 2 square root of x minus a.

03:15 And now we can simplify a little bit this expression.

03:18 This is, we can take a common denominator, this expression to square root of x minus a.

03:27 We have a 1 here.

03:30 So we get 2 times square root of x minus a times square root of x minus a.

03:36 So we get 2x minus a because it's 2 square root of x minus a, because it's 2 square root of x minus a square.

03:42 And so we get this plus x and so this is 2x minus 2a plus x over 2 square root of x minus a and we end up with this expression 3x minus 2a over 2 square root of x minus a and that's the final formula for the derivative of f respect to x, 3x minus 2a divided by 2 times square root of x minus a.

04:20 And now the critical points corresponds to this derivative being equal to 0.

04:30 So the derivative of f is 0 is the same as the numerator being 0, but at that value of the x that nullifies the numerator, we cannot have a 0 in the denominator.

04:46 That's very important to remark.

04:49 So this is zero.

04:51 Whenever the x we get from this equation is not a value that nullifies the denominator.

04:58 So this is equivalent to x being equal to 2a over 3.

05:04 So that's the x coordinate of a point.

05:09 So there's only one point where the derivative can be zero.

05:16 Okay, so we have that.

05:21 And now we cannot say directly or without any confirmation that this is a critical point for every value of a.

05:35 Because we have a square root of x minus a in the denominator that could maybe be not defined.

05:43 So we get to examine when that value of x can be put in the denominator here.

05:50 So, and we have another thing to remark here is to say that even though the domain of the function f is all values of x for which x is greater than or equal to a, in the case of this derivative, the domain of this derivative is similar to that, the domain of f, but the value of a itself is not included because we cannot put x equal a in this denominator because we because we get a zero.

06:25 So in general then the condition on this value we got here to be one value that can be put in the formula of the derivative is that this value of x is greater than a.

06:45 In fact, cannot be equal to a.

06:47 So we must have that 2a over 3b greater than a in order to have that a derivative at 2a over 3 is well defined.

07:22 Okay, so this inequality we get to verify.

07:27 So 2a over 3 greater than a, we multiply by 3...

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