Find extrema on CLOSED interval!

Extreme Value Theorem

If f is continuous on a [closed] interval, then f has an ABSOLUTE MAXIMUM & MINIMUM

wait, there is a max and min even when it is a straight line?

like: y=1

YES!!! IF it is defined on a closed interval

the max and min of y=1 is 1!

f(c) is the max if f(c) >= f(x) for all x values

f(c) is the min if f(c) <= f(x) for all x values

1. all x values means it is continuous

How do you find the Absolute maximum and minimum?

Candidates test :  test all points where critical points can exist to see which values are the highest(maximum)                                  

                           and the lowest(minimum)

Candidates will be endpoints and critical points, explained soon below!

Critical point- point on f(x) where the slope is equal to zero ( which means the derivative is zero )  or point on f(x) where the slope is undefined ( the derivative does not exist )

STEPS TO SOLVING:

1. find derivative(the 1st derivative!) of the function
2. find critical points (these will be candidate)

• critical points: 
• solve for the zeros(roots, x-intercepts) of the DERIVATIVE  because this will make the entire derivative equal zero
• solve for the points which make the function undefined ( check for denominators with variables and square root functions-- you can not take the sqrt of a negative number!)

     3.  Critical points and the ENDPOINTS of the interval are your CANDIDATES

     4. PERFORM CANDIDATE TEST 

• candidate test:
• identify candidates (Critical points and ENDPOINTS of interval)
• plug in each candidate into Original function
• the highest value found = MAXIMUM
• lowest value found = MINIMUM
• REMEMBER, if the function is continuous on a closed interval, THERE HAS to be a maximum and minimum!

Example 1:

f(x) = 1/3X³ - 4x² + 12x -5 at domain [0,9]

what is the absolute maximum of this function?

1. find the derivative

                        f(x)=  x^2 -8x +12
2. critical points exist where this derivative is zero/undefined

f(x) is not ever undefined b/c continuous at all points 

so, we will just set this derivative equal to zero

                          0 = x^2 - 8x +12
                          0 = (x-6) (x-2)

x is zero 6 and 2.

3. identify candidates(endpoints and critical values)
       candidates are 6, 2, 0, 9
4. plug in these candidates into ORIGINAL FUNCTION, the highest number is the maximum

THE ANSWER IS 9!