Suppose that f(x) = (x + 6)(x - 5)^2. (A) Find all critical...

Suppose that f(x) = (x + 6)(x - 5)^2. (A) Find all critical values of f. If there are no critical values, enter -1000. If there are more than one, enter them separated by commas. Critical value(s) = (B) Use the second derivative test to find the x-coordinates of all local maxima of f. If there are no local maxima, enter -1000. If there are more than one, enter them separated by commas. Local maxima at x = (C) Use the second derivative test to find the x-coordinates of all local minima of f. If there are no local minima, enter -1000. If there are more than one, enter them separated by commas. Local minima at x =

Transcript

00:01 The critical values occur where the derivative of the function is equal to zero.

00:08 So let's start by finding the derivative of the function.

00:11 So we're differentiating with respect to x.

00:13 I'm going to use the product rule.

00:16 So first let's hold x plus 6 constant and we'll differentiate x minus 5 squared.

00:22 So using the power rule we get 2 times x minus 5.

00:26 Now we'll differentiate x plus 6 which is just 1 and we multiply by x minus 5 squared.

00:36 Okay let's simplify this.

00:38 I'm going to factor out x minus 5.

00:41 So then we're left with 2 times x plus 6 plus x minus 5.

00:53 Okay so we can simplify this x minus 5 multiplied by 2x plus x is 3x and 2 times 6 is 12 minus 5 is 7.

01:08 Okay so to find the critical values we'll set this equal to 0 and solve for x.

01:14 So i see that we have two solutions.

01:18 The first solution is 0 equals x minus 5 which tells us that x is equal to 5.

01:26 The second solution is 0 equals 3x plus 7 which tells me that negative 7 equals 3x.

01:36 So x is equal to negative 7 thirds.

01:40 So these are the two critical values of f.

01:47 Let's call these critical values xc.

01:51 Okay so now we'll use a second derivative test.

01:55 The second derivative test says that if the second derivative of f at the critical value is positive, okay so greater than zero, then the function evaluated at that critical point is a local minimum.

02:16 Okay and if the second derivative evaluated at a critical value is negative, so less than zero, then the value of this function at that critical point is a local maximum.

02:31 Okay this is because if the second derivative is positive that means that the function is concave up...

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