- WE WANT TO DETERMINE 
ANY RELATIVE EXTREMA
USING THE FIRST DERIVATIVE 
TEST.
SO THE FIRST STEP IS TO 
DETERMINE THE POSSIBLE LOCATIONS
OF RELATIVE MAX AND MINS,
AND THESE OCCUR AT THE CRITICAL 
NUMBERS,
WHICH IS WHERE THE FIRST 
DERIVATIVE
WOULD BE EQUAL TO ZERO OR WHERE 
THE DERIVATIVE IS UNDEFINED.
SO WE'LL START BY DETERMINING 
THE DERIVATIVE
OF THE GIVEN FUNCTION.
F PRIME OF X IS GOING TO BE 
EQUAL TO 12X SQUARED - 30X - 18,
AND THE DERIVATIVE OF TWELVE 
WOULD BE ZERO.
SO THIS DERIVATIVE FUNCTION 
IS NEVER UNDEFINED.
SO NOW, WE WANT TO DETERMINE 
WHERE IT'S EQUAL TO ZERO.
SO WE'LL SET IT EQUAL TO ZERO 
AND SOLVE.
WE HAVE A QUADRATIC EQUATION.
SO LET'S SEE IF WE CAN FACTOR 
THIS.
THE GREATEST COMMON FACTOR OF 
THESE THREE TERMS WOULD BE SIX,
LEAVING US WITH 
2X SQUARED - 5X - 3.
SO LOOKS LIKE THIS WILL FACTOR.
THE FIRST POSITIONS OF THE 
BINOMIAL FACTORS
MUST COME FROM THE FACTORS 
OF 2X SQUARED.
THAT WOULD BE 2X AND X,
AND THE SECOND POSITIONS MUST 
COME FROM THE FACTORS OF -3
SO THAT THE INNER PRODUCT AND 
OUTER PRODUCT HAS A SUM OF -5X.
SO IF WE PUT MINUS 
THREE HERE AND PLUS ONE HERE,
NOTICE HOW WE HAVE -6X + 1X, 
AND THAT DOES GIVE US -5X.
SO THE DERIVATIVE IS EQUAL TO 
ZERO
WHEN 2X + 1 = 0 
OR WHEN X - 3 = 0.
SO SOLVING FOR X, WE WOULD 
SUBTRACT ONE AND DIVIDE BY TWO.
WE HAVE X EQUALS -1/2 HERE, 
AND WE HAVE X EQUALS THREE HERE.
AND THESE TWO X VALUES 
ARE THE CRITICAL NUMBERS,
WHICH ARE THE POSSIBLE LOCATIONS 
FOR ANY RELATIVE EXTREMA.
SO NOW, WHAT WE'RE GOING TO DO
IS TAKE THE DOMAIN 
OF THE GIVEN FUNCTION,
WHICH IS ALL REAL NUMBERS AND 
DIVIDE IT UP INTO INTERVALS
USING THESE CRITICAL NUMBERS.
SO WHAT I MEAN BY THAT 
IS IF WE HAVE A NUMBER LINE
THAT REPRESENTS ALL REAL 
NUMBERS,
AND THEN WE PLOT -1/2, 
AND THEN WE PLOT THREE,
NOTICE HOW WE HAVE THREE 
INTERVALS:
THE INTERVAL ON THE LEFT, 
IN THE MIDDLE AND ON THE RIGHT.
AND THEN WE'RE GOING TO TEST THE 
SIGN OF THE FIRST DERIVATIVE
IN EACH INTERVAL, WHICH WILL 
TELL US WHETHER THE FUNCTION
IS INCREASING OR DECREASING 
IN THE INTERVAL.
AND THEN FROM THERE, WE'RE GOING 
TO DETERMINE
IF WE HAVE ANY RELATIVE MAX OR 
MINS.
SO I THINK IT'S OFTEN BEST 
TO ORGANIZE THIS INFORMATION
IN A TABLE, AND I'VE ALREADY 
DONE THAT TO SAVE SOME TIME.
NOTICE THE FIRST INTERVAL 
IS FROM -8 TO -1/2,
AND THEN WE HAVE 
FROM -1/2 TO THREE,
AND THEN FROM THREE TO INFINITY.
SO WE'RE GOING TO PICK A TEST 
VALUE IN EACH INTERVAL,
AND THEN USING THE TEST VALUE,
WE'LL DETERMINE THE SIGN 
OF THE FIRST DERIVATIVE.
AND THEN FROM THERE, 
WE CAN DRAW OUR CONCLUSIONS
AND THEN DETERMINE IF WE HAVE 
ANY RELATIVE EXTREMA.
SO LET'S START BY DETERMINING 
A POSSIBLE TEST VALUE.
WE NEED SOME NUMBER 
THAT WOULD BE IN THIS INTERVAL;
FOR EXAMPLE, -1.
IN THIS INTERVAL WE COULD SELECT 
ZERO,
AND HERE WE COULD SELECT FOUR.
NOW, WE'LL SUB THESE X VALUES 
INTO THE FIRST DERIVATIVE,
WHICH I'VE COPIED HERE FROM 
THE PREVIOUS SCREEN.
IT DOESN'T MATTER WHETHER WE USE 
THE FACTORED FORM
OR THE EXPANDED FORM 
OF THE FIRST DERIVATIVE.
I'M GOING TO GO AHEAD 
AND USE THE FACTORED FORM.
SO F PRIME OF -1 WOULD BE 6 x 2 
x -1 + 1.
THAT'S GOING TO BE -1 x -1 - 3.
THAT'S -4, AND THIS IS GOING 
TO BE EQUAL TO 24,
WHICH IS POSITIVE.
SO THE SIGN OF THE FIRST 
DERIVATIVE IS POSITIVE
IN THIS ENTIRE INTERVAL,
AND NOW, WE DETERMINE F PRIME
OF ZERO FOR THE MIDDLE INTERVAL.
SO WE'LL HAVE 6 x 2 x 0 + 1 = 1, 
AND 0 - 3 = -3.
SO THIS IS -18, 
WHICH IS NEGATIVE,
AND THEN FOR THE LAST INTERVAL 
WE'LL TEST THE VALUE OF X = 4.
SO WE HAVE 6 x 2 x 4 + 1.
THAT'S 9 AND HERE WE HAVE 
4 - 3; THAT'S 1.
SO WE HAVE 6 x 9, 
WHICH IS EQUAL TO 54,
WHICH IS POSITIVE.
NOW, THE SIGNS OF THE DERIVATIVE 
TELL US WHETHER THE FUNCTION
IS INCREASING OR DECREASING
ON THAT INTERVAL.
SO ON THIS INTERVAL, SINCE THE 
DERIVATIVE IS POSITIVE,
THE FUNCTION IS INCREASING.
ON THIS INTERVAL, THE FIRST 
DERIVATIVE IS NEGATIVE.
SO IT'S DECREASING, 
AND THEN IT'S POSITIVE AGAIN.
SO THE FUNCTION 
IS INCREASING ON THIS INTERVAL.
SO THAT TELLS US THE FUNCTION.
SO TO VISUALIZE THIS,
THE FUNCTION IS INCREASING 
ON THIS INTERVAL,
DECREASING 
ON THIS MIDDLE INTERVAL
AND THEN INCREASING AGAIN ON 
THIS LAST INTERVAL.
SO AT X = -1/2 WE HAVE A HIGH 
POINT OR A RELATIVE MAXIMUM,
AND AT X = 3 WE HAVE A LOW POINT 
OR RELATIVE MINIMUM.
SO LET'S GO AHEAD 
AND RECORD THIS DOWN HERE.
WE HAVE A RELATIVE MAXIMUM 
AT X = -1/2,
AND WE HAVE A RELATIVE MINIMUM 
AT X = 3.
SO WE KNOW THE LOCATIONS, BUT WE 
ACTUALLY HAVEN'T DETERMINED
THE RELATIVE MAX AND MIN VALUES 
YET.
AND THOSE ARE THE FUNCTION 
VALUES OF THE ORIGINAL FUNCTION.
SO WE'LL DETERMINE F OF -1/2 TO 
DETERMINE THE RELATIVE MAXIMUM,
AND THEN WE'LL DETERMINE F 
OF THREE
TO DETERMINE THE RELATIVE 
MINIMUM.
TO DO THAT, WE'LL 
GO AHEAD AND USE THE CALCULATOR.
SO I'VE ALREADY TYPED THE 
ORIGINAL FUNCTION INTO Y1,
AND NOW, WE'RE GOING TO GO BACK 
TO THE HOME SCREEN.
WE'RE GOING TO PRESS VARS, 
RIGHT ARROW, ENTER, ENTER
TO SELECT Y1, AND THEN WE CAN 
JUST TYPE IN -1/2,
AND A SET OF PARENTHESES.
SO THE RELATIVE MAXIMUM 
IS 16.75 OR 16 3/4.
SO THEN WE'LL PRESS VARS, RIGHT 
ARROW, ENTER, ENTER,
AND THEN WE'LL ENTER THREE 
AND A SET OF PARENTHESES.
SO OUR RELATIVE MINIMUM 
IS EQUAL TO -69.
SO IT'S IMPORTANT TO REMEMBER
THAT THE RELATIVE MAX AND MIN 
VALUES ARE FUNCTION VALUES,
AND THE CRITICAL NUMBERS 
ARE THE LOCATIONS.
SO WE HAD A MAX OF 16.75 
AND A MIN OF -69.
NOW, I DO WANT TO MENTION 
THAT SOMETIMES
WE'LL SEE TEXTBOOKS LIST THE 
RELATIVE EXTREMA
AS ORDERED PAIRS, WHICH IS OKAY.
THIS WOULD BE THE POINT 
(-1/2, 16.75)
AS THE RELATIVE MAX, AND THIS 
WOULD BE THE POINT (3, -69)
FOR THE RELATIVE MINIMUM,
BUT I THINK IT'S IMPORTANT 
TO REMEMBER
THAT THE FUNCTION VALUE
IS ACTUALLY THE RELATIVE 
MAX OR MIN,
AND THE X VALUE IS THE LOCATION.
AND LET'S GO AHEAD AND VERIFY 
THIS GRAPHICALLY.
HERE'S THE RELATIVE MAX 
WE FOUND,
AND HERE'S THE RELATIVE MIN 
THAT WE FOUND.
THIS WAS (-1/2, 16 3/4), 
AND THIS WAS (3, -69).
I HOPE YOU FOUND THIS HELPFUL.
