Scribe Post 11/14: Extrema

Hey guys!
**To study, try to do the previous problems on the blank test, which are on the iCal
Remember, problems that were multiple choice before probably won’t be in multiple choice form on the test! Some of these will be the type of problem where you just replace values, but others will not, so make sure you actually know how to do everything!

Now, on to the real mathy stuff.

We started by talking about absolute (global) extreme. The definition can be found on page 191 in our textbooks, but here it is anyway!

• Let f be a function with domain D. Then, f(c) is the

• a) Absolute maximum value on D if and only if f(x) is less than or equal to f(c) for all values of x in D
• b) Absolute minimum value on D if and only if f(x) is greater than or equal to f(c) for all values of x in D
• These are the largest or smallest y values!

Example:
If we look at the function f(x)=x2 on various domains, we realize that there are different maximums and minimums depending on the restrictions. When the domain is all reals, there is no max because the function is continuously increasing, but there is a minimum of zero

• On domain [0,2], there is a max of 4 and a minimum of zero, because it is a closed interval
• On domain (0,2], there is no min because it is an open interval on that side and the max is 4, because that side is still closed
• On domain (0,2) there is no min or max, because both sides are open

 This leads us to our next theorem, the Extreme Value Theorem, which states that:

• If f is continuous on a closed interval [a,b], then there is both an absolute (global) max and an absolute (global) min

Local Extreme Values: The definition can be found on page 193 in our textbooks. It is as follows:
Let c be a point of the domain of the function f. Then, f(c) is either

• a) a local maximum value: at c if and only f f(x) is less than or equal to f(c) for all values x in some open interval containing c
• b) a local minimum value of c, if and only if f(x) is greater than or equal to f(c) for all values x in some open interval containing c

• In other words; a function f has a local max or min at an endpoint c if the appropriate inequality holds for all x in some half open domain interval containing c

We also took some time to review an old term: critical points. Our new definition states that they are values of x where f’ is zero or undefined. More specifically, the points where f’ is zero are called stationary points, because it is where the motion briefly stops.  The local extreme are always either found at endpoints or critical points.

•  One of the absolute max or mins has to be one of the local max or mins!!
• The max/min will be either an endpoint or at a point of change, such as a cusp or corner, where the derivative is undefined; or when the derivative is zero, which can be caused by a gradual change.

**Just remember to be careful, because this can actually be a little tricky. For example, f(x)=x³ has a derivative of zero when evaluated at zero, however, it is not a max or a min. It’s a lot like the argument of a square being a rectangle, but a rectangle isn’t a square kind of thing.




After learning all of the definitions, we put them to practice. We were told to find the extreme values of f(x)=x^2/3 on [-2,3] algebraically and confirm graphically.

• First, find the derivative (you can use the power rule in this case!) 
• Then, decide on the critical points. Include the end points and the value that makes the denominator undefined (-2,0,3). In this case, the derivative can never equal zero! Remember, these are x values!
• It can be helpful to make a chart of these values. Use it to plug the x values back into the original equation. From there, just take the smallest value produced for the minimum and the largest value for the maximum! Tada!
  
  
  What if we were asked to find the extreme values of a more difficult function? Well, we can do it!
  
  
• We find the derivative by using the chain rule, shown above
• Then, identify the critical points, which are x=0,2,-2
• Plug into original function and pick out largest and smallest values
• We then can see that the function has two undefined values. Therefore, there is no max, but there is a minimum of 1/2.