Absolute maxima and minima Determine the location and value of the absolute extreme values of f

Absolute maxima and minima Determine the location and value...

Key Concepts

Extreme Value Theorem

The Extreme Value Theorem guarantees that if a function is continuous on a closed interval, then it must have both an absolute maximum and an absolute minimum within that interval. This theorem provides the foundation for searching for these extreme values.

Critical Points

Critical points are points in the domain of a function where the derivative is zero or undefined. These points are important candidates for local extrema and, when considered along with endpoint evaluations, help in determining the absolute extreme values on a closed interval.

Absolute Extrema

Absolute extrema refer to the highest (maximum) and lowest (minimum) values that a function attains over a specified domain. For functions defined on closed intervals, these extrema can occur at interior points or at the endpoints.

Endpoint Evaluation

Evaluating the function at the endpoints of the interval is necessary because absolute extrema may occur at these boundary points. For continuous functions on closed intervals, the endpoints must be checked alongside any critical points to ensure that all potential extreme values are identified.

Transcript

00:01 Alright, let's talk about how to find the extreme values of this function.

00:03 So in order to find the extreme values, the very first thing you need is to find all the critical points first.

00:09 So find the critical points by taking the first derivative and then setting that derivative equal to zero.

00:15 Okay? so our first derivative is going to be f of x, f prime of x, is equal to 4x cubed minus 12x squared plus 8x squared plus 8.

00:35 All right now take our first derivative and set it equal to zero all right so zero equals 4x cubed minus 12x squared plus 8x factor what we have in common which is a 4x so take out of 4x and we're left with x squared minus 3x plus 2 okay now we can use 0 property rule to separate these two.

01:11 We get 4x is equal to 0, and x squared minus 3x plus 2 is equal to 0.

01:19 Divide by 4, and one of our critical values is going to be 0.

01:25 Now we need to factor this function over here.

01:28 So we're going to have, we need the factors of 2 that'll give us negative 3.

01:35 So obviously the only factors of 2 are 1 and 2.

01:38 And if we need a negative 3, that means we're going to have a negative 1 as well as a negative 2.

01:46 Now, these are equal to 0.

01:48 Using our 0 property rule, again, we have that x minus 1 is equal to 0, x minus 2 is equal to 0.

01:57 So therefore, x is equal to 1 and x is equal to 2.

02:07 So these are our three critical values...