Suppose that f(x) = (x^2 + 11)(9 - x^2). (A) Find all...

Suppose that f(x) = (x^2 + 11)(9 - x^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) = 0 (B) Use interval notation to indicate where f(x) is increasing. Note: When using interval notation in WeBWork, you use I for $\infty$, I for $-\infty$, and U for the union symbol. If there are no values that satisfy the required condition, then enter "{}" without the quotation marks. Increasing: (-1, 0) (C) Use interval notation to indicate where f(x) is decreasing. Decreasing: (0, 1) (D) 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 = 0 (E) 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 = -1000 (F) Use interval notation to indicate where f(x) is concave up. Concave up: 0

Transcript

00:01 We have the function f of x equals x squared minus four and now let's raise the six power.

00:06 We want to find the critical values, the intervals where the function is increasing and decreasing, local minima, maxima, and local minima, and then the intervals where the function is concave up and concave down.

00:16 So first of all, we can just get a general shape of this graph.

00:23 Um, first of all, it's worth noting that f of x is always going to be, in this case, since we're raising this to the sixth power, that will always be greater than or equal to zero.

00:42 Since we're taking this value and raising the sixth power, since that degree is even, anything to the sixth power, if we have u to the sixth power, where in this case, um, u is the x squared minus four, u to the sixth power will always be greater than or equal to zero.

01:01 Anything to an even exponent will be equal to zero or greater than zero.

01:05 So that means the function will always be positive, so it's always going to be above the x -axis, and then we can, um, find the zeros of that function.

01:17 If we take x squared minus four to the sixth power, instead of equal to zero, we can solve for x.

01:22 We can take the sixth root of both sides, and that gives us x squared minus four equals the sixth root of zero, which is zero.

01:31 And then we can find the two zeros of that by factoring that into x minus two and x plus two equals zero.

01:39 That means the two solutions to that equation will be x equals two and x equals negative two.

01:44 So that means negative two and two would be zeros on the graph, the only two zeros of that graph, and if we wanted to find the, um, y -intercept or f of zero, that would be zero squared minus four to the sixth power, which would just be negative four to the sixth power, which would be, um, 4 ,096.

02:09 So a really big, um, y -intercept, that's at 4 ,096.

02:14 So the graph we know has to pass through these three points.

02:18 Um, so if we want to find the critical values now before we sketch this graph, that would be when the derivative of this function equals zero.

02:30 So let's find the derivative first, f prime of x.

02:33 We would do that by using the power rule.

02:36 Um, we'd bring that six down to the front, six times x squared minus four, and that would go down to the fifth power, and we multiply by the derivative of two x minus, or x squared minus four, which would just be two x.

02:50 The derivative of x squared is two x.

02:53 So we could simplify that a bit and get six times two x is 12 x times x squared minus four to the fifth power.

03:03 Okay, um, the critical points would happen when this function is equal to zero.

03:09 So let's set each factor equal to zero.

03:11 If we have 12 x equals zero, that means x equals zero would be one critical point.

03:17 So x equals zero is one critical value, and if we take the other factor, x squared minus four to the fifth power equals zero, we could take the fifth through to both sides, and that would give us x squared minus four equals zero, and we've already solved that equation once.

03:34 That's x equals two and x equals negative two.

03:38 So the other critical values are x equals positive two and x equals negative two, and those are the values we've already really focused on, um, when we graphed this function.

03:53 The critical values occur at zero, two, and negative two.

03:56 So, um, if we want to, uh, knowing that this function is always positive, and to go through these three points, the graph must look like this right here.

04:11 Um, the critical values are the, also note, we can look at that as turning points.

04:15 Those are the only times where that function will change direction.

04:18 So now we can easily see where this function is increasing.

04:21 It's increasing on, um, this interval here and this interval here.

04:28 That would be from negative two to zero, it's increasing, and then we'd say union symbol, and then the other interval is from two to infinity is when it's increasing, and the function is decreasing on this interval as it goes down from left to right.

04:45 So that would be from, um, negative infinity to negative two, and then the other interval would be from zero to two.

04:55 Local maxima would happen right here at that point, and we're just going to give the x coordinate of that point.

05:02 That's when x is zero, we have, uh, local maxima...

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