Find the absolute extrema if they exist, as well as all...

Find the absolute extrema if they exist, as well as all values of x where they occur, for each function, and specified domain. If you have one, use a graphing calculator to verify your answers.
$$f(x)=\frac{(4+x)}{(3-x)} ;[-3,0]$$

Key Concepts

Extreme Value Theorem

The Extreme Value Theorem states that if a function is continuous on a closed interval, then it must have both a maximum and a minimum value on that interval. This is a fundamental result in calculus that guarantees the existence of absolute extrema under the right conditions.

Critical Points

Critical points are points in the interior of the domain where the derivative is zero or does not exist. These points are important because they are potential candidates for local extrema, and when working with closed intervals, they must be checked along with the endpoints to find absolute extrema.

Closed Interval Method

The Closed Interval Method is a procedure used to find absolute extrema on a closed interval. It involves finding all the critical points inside the interval, evaluating the function at these points and at the endpoints of the interval, and then comparing the values to determine the absolute maximum and minimum.

Endpoint Evaluation

Evaluating the function at the endpoints of the interval is crucial because, according to the Extreme Value Theorem, absolute extrema can occur at the boundaries of the domain, not just at critical points. This step ensures that no extreme value is overlooked.

Transcript

00:01 Okay, we're going to go ahead and walk through the process of how to determine absolute extrema for a function.

00:13 And the function we're going to do it for is f of x is equal to 4 plus x divided by 3 minus x.

00:25 And we're going to do it for the close interval from negative 3 to 0.

00:30 And so the very first thing that we have to do is to take the derivative of the function.

00:39 And so we're going to have to use the quotient rule.

00:43 So the derivative of the numerator, which is one times the denominator, minus the numerator times the derivative of the denominator all over the denominator squared.

01:04 And so we're going to simplify this, and we get 3 minus x.

01:12 This subtraction and this negative make a positive.

01:16 So this would be plus 4 plus x all over 3 minus x squared, which is 7 over 3 minus x squared.

01:29 So that's the first thing is to take the derivative.

01:32 The second thing is to find what are called critical values, and you do them two ways.

01:39 You set your derivative equal to zero, and you set your derivative equal to undefined.

01:48 And the undefined part will only occur if you have x values or x variable in the denominator, which we do.

02:00 But f prime of x equal to zero will only be if i have x variable in the numerator, which i don't have in this derivative.

02:15 So i will have no critical values based off of f prime of x equal to zero, but i will have critical values for f prime of x equal to undefined.

02:27 And this undefined means that my denominator actually equals zero.

02:32 So this means my denominator equals zero.

02:36 So i will set my denominator equal to zero and solve for that x value.

02:43 And that x value is going to occur when x is equal to 3.

02:48 So this is the only critical value i have...

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