- WELCOME TO A VIDEO THAT WILL DESCRIBE THE KEY FEATURES
OF A FUNCTION USING CALCULUS.
MORE SPECIFICALLY, GIVEN A FUNCTION
WE'LL DETERMINE THE INTERVALS
FOR WHICH THE FUNCTION IS INCREASING OR DECREASING
AND WHERE IT'S CONCAVE UP OR DOWN
AS WELL AS STATE ANY RELATIVE EXTREMA
AND POINTS OF INFLECTION.
SO HERE'S OUR FUNCTION.
THE FIRST THING WE NEED TO DO IS DETERMINE
THE CRITICAL NUMBERS.
WE DO THAT BY FINDING THE FIRST DERIVATIVE
AND DETERMINING WHERE THE FIRST DERIVATIVE
IS UNDEFINED OR EQUAL TO ZERO.
SO APPLYING THE POWER RULE HERE
WE'LL HAVE F PRIME OF X = -12X CUBED + 72X SQUARED - 96X.
THIS WILL NEVER BE UNDEFINED SO LET'S GO AHEAD
AND SET IT EQUAL TO ZERO AND SOLVE.
THE GREATEST COMMON FACTOR WOULD BE -12X
LEAVING US WITH X SQUARED - 6X + 8.
NOW THIS TRINOMIAL FACTORS.
THE FACTORS OF 8 THAT ADD TO -6 THAT WOULD BE A -4 AND A -2.
SO THE CRITICAL NUMBERS ARE X = 0 FROM THIS FACTOR.
X = 4 AND X = 2.
THE DOMAIN OF THIS FUNCTION IS ALL REAL
SO WE'RE GOING TO DIVIDE THE REAL NUMBER LINE
INTO INTERVALS USING THESE CRITICAL NUMBERS AS DIVIDERS.
SO HERE WE'LL HAVE OUR INTERVALS
THEN WE'LL HAVE THE SIGN OF THE FIRST DERIVATIVE.
THAT WILL TELL US WHETHER IT'S INCREASING OR DECREASING
AND THEN WE'LL DRAW OUR CONCLUSION.
SINCE THE CRITICAL NUMBERS WERE 0, 2, 4,
AND THE DOMAIN IS ALL REAL'S
THE FIRST INTERVAL WILL BE FROM NEGATIVE INFINITY TO ZERO
AND THEN WE'LL HAVE THE INTERVAL FROM 0 TO 2, 2 TO 4,
AND 4 TO INFINITY.
NOW WE'LL PICK A TEST VALUE IN EACH INTERVAL
TO DETERMINE THE SIGN OF THE FIRST DERIVATIVE.
AND WE'RE GOING TO GO AHEAD AND USE THE CALCULATOR FOR THIS.
LET'S GO AHEAD AND TYPE THE ORIGINAL FUNCTION IN Y1
AND THEN WE'LL TYPE THE DERIVATIVE INTO Y2.
NOW WE'LL USE THE TABLE FEATURE TO DETERMINE
THE SIGN OF THE FIRST DERIVATIVE IN EACH OF THESE INTERVALS.
SO WE'RE GOING TO PRESS 2nd GRAPH
AND WE'RE CHECKING THE SIGN OF THE FIRST DERIVATIVE
SO WE'RE LOOKING AT THE Y VALUES IN Y2.
SO FOR THIS FIRST INTERVAL LET'S GO AHEAD AND SELECT -1
AS OUR TEST VALUE.
NOTICE THE FIRST DERIVATIVE IS POSITIVE
AND THEN FOR THIS INTERVAL WE'LL USE THE VALUE OF 1.
NOTICE THE FIRST DERIVATIVE IS NEGATIVE
AND THEN FROM THIS INTERVAL FROM 2 TO 4
WE'LL SELECT X = 3 AND IT'S POSITIVE.
AND THEN THE LAST INTERVAL WAS FROM 4 TO INFINITY.
LET'S GO AHEAD AND USE 5 AND NOTICE THAT IT'S NEGATIVE.
SO IT DOES ALTERNATE FROM POSITIVE TO NEGATIVE
IN EACH INTERVAL.
LET'S GO AHEAD AND RECORD THAT.
IT WAS POSITIVE, NEGATIVE, POSITIVE, AND NEGATIVE.
SO THAT TELLS US ON THIS INTERVAL
THE FUNCTION IS INCREASING, ON THIS INTERVAL IT'S DECREASING,
ON THIS INTERVAL IT'S INCREASING,
AND THIS INTERVAL IT'S DECREASING.
SO AGAIN, WE KNOW THE FUNCTION
IS GOING UP AND THEN DOWN AND THEN UP AND THEN DOWN.
WELL IF THE FUNCTION CHANGES FROM INCREASING TO DECREASING
WE WOULD HAVE A RELATIVE MAXIMUM AT X = 0
AND THEN IT CHANGES FROM DECREASING TO INCREASING
AT X = 2.
SO WE HAVE A RELATIVE MINIMUM HERE AND THEN AT 4
IT CHANGES FROM INCREASING TO DECREASING AGAIN
SO WE HAVE ANOTHER RELATIVE MAXIMUM.
SO AGAIN, WE SAID WE HAD A RELATIVE MAX AT X = 0
AS WELL AS X = 4 AND THEN WE HAD A RELATIVE MINIMUM AT X = 2.
THE VALUE OF THE RELATIVE EXTREMA IS THE FUNCTION VALUE
OR THE Y VALUE SO LET'S GO AHEAD AND FIND THAT AS WELL.
SO WHEN WE GO BACK TO THE CALCULATOR,
NOW WE'RE GOING TO BE LOOKING AT Y1 FOR THE FUNCTION VALUE.
WE'LL PRESS 2nd GRAPH,
X = 0 THE FUNCTION VALUE IS ZERO
AND NOTICE THAT 4 IT'S ALSO EQUAL TO ZERO AND THEN AT X = 2
THE FUNCTION VALUE IS -48.
SO THE RELATIVE MAXIMUM ARE BOTH ZERO
AND THE RELATIVE MINIMUM WAS -48.
SO NOW WE'VE DETERMINED WHERE THE FUNCTION
IS INCREASING OR DECREASING AS WELL AS RELATIVE EXTREMA.
NOW LET'S TALK ABOUT CONCAVITY AND THE 2nd DERIVATIVE.
TO DETERMINE THE POSSIBLE POINTS OF INFLECTION
WE NEED TO FIND THE 2nd DERIVATIVE AND DETERMINE
WHERE THAT IS UNDEFINED OR EQUAL TO ZERO.
HERE'S THE 1st DERIVATIVE
LET'S GO AHEAD AND FIND THE 2nd DERIVATIVE.
THIS WOULD BE -36X SQUARED + 144X - 96.
THIS WILL NEVER BE UNDEFINED
SO LET'S GO AHEAD AND SET IT EQUAL TO ZERO AND SOLVE.
THE GREATEST COMMON FACTOR HERE WOULD BE -12
SO WE'LL HAVE 3X SQUARED - 12X + 8.
NOW THIS TRINOMIAL IS NOT FACTORABLE
SO WE'RE GOING TO HAVE TO GO AHEAD
AND USE THE QUADRATIC FORMULA
WHERE "A" = 3, B = -12, AND C = 8.
SO WE'LL HAVE X = -B + OR - THE SQUARE ROOT OF B SQUARED - 4
x "A" x C ALL OVER 2 x "A."
SO HERE WE'LL HAVE 12 + OR - THE SQUARE ROOT OF--
THIS IS GOING TO BE 144 - 96, THAT'S GOING TO GIVE US 48/6.
NOW 48 IS 16 x 3 SO THAT WOULD BE 12
+ OR - THE SQUARE ROOT OF 16 WOULD BE 4 SQUARE ROOT 3
AND I ALWAYS LIKE TO SIMPLIFY THIS BY WRITING IT
AS TWO DIFFERENT FRACTIONS.
SO WE HAVE 12/6 + OR - 4 SQUARE ROOT 3/6.
SO THIS IS = 2 + OR - THIS WOULD BE 2 SQUARE ROOT 3/3.
SO THESE X VALUES ARE THE LOCATIONS
OF POSSIBLE POINTS OF INFLECTION.
FOR THE PURPOSE OF OUR TABLE,
IT MIGHT BE HELPFUL TO CONVERT THESE TO DECIMALS.
2 + 2 SQUARE ROOT 3 DIVIDED BY 3 IS APPROXIMATELY 3.155
AND 2 - 2 SQUARE ROOT 3/3 IS APPROXIMATELY 0.845.
SO USING THESE X VALUES WE'LL DIVIDE UP THE DOMAIN
INTO 3 DIFFERENT INTERVALS.
THE FIRST INTERVAL WILL BE FROM NEGATIVE INFINITY TO .845.
THE SECOND INTERVAL WILL BE FROM .845 TO 3.155
AND THEN WE'LL HAVE THE INTERVAL FROM 3.155 TO POSITIVE INFINITY.
NOW WE'RE GOING TO TEST THE SIGN IN OF 2nd DERIVATIVE
IN EACH OF THESE INTERVALS.
SO LET'S GO AHEAD AND GO BACK TO OUR CALCULATOR.
I'VE ALREADY TYPED THE 2nd DERIVATIVE INTO Y3
SO NOW WE'LL PICK A TEST VALUE IN EACH OF THESE INTERVALS
AND DETERMINE THE SIGN IN Y3.
SO WE'LL PRESS 2nd GRAPH, -1 IS IN THIS FIRST INTERVAL
SO LET'S GO OVER TO Y3.
NOTICE THAT IT'S NEGATIVE.
WE COULD USE 1 FOR THE SECOND INTERVAL
AND NOTICE THAT IT'S POSITIVE AND THE LAST INTERVAL
WE COULD USE X = 4 NOTICE THAT IT'S NEGATIVE.
SO IT'S NEGATIVE, POSITIVE, AND THEN NEGATIVE.
WELL IF THE 2nd DERIVATIVE IS NEGATIVE ON THIS INTERVAL
WE KNOW THAT THE FUNCTION WOULD BE CONCAVE DOWN.
IF IT'S POSITIVE IN THIS INTERVAL IT'S CONCAVE UP
AND THIS INTERVAL IT WOULD BE CONCAVE DOWN.
SINCE IT DOES CHANGE CONCAVITY AT .845 AND 3.155
WE DO HAVE TWO POINTS OF INFLECTION.
LET'S GO AHEAD AND FIND THE Y COORDINATES
OF THOSE POINTS OF INFLECTION.
ONE THING WE SHOULD BE AWARE OF THAT
IF THESE X VALUES HAVE BEEN ROUNDED.
THIS ACTUALLY CAME FROM 2 - 2 SQUARE ROOT 3/3
AND THIS X VALUE CAME FROM ROUNDING 2 + 2 SQUARE ROOT 3/3.
SO IF WE WANT TO BE MORE PRECISE IN FINDING
THE Y COORDINATES OF THESE POINTS OF INFLECTION
WE SHOULD USE THE EXACT VALUES.
LET'S GO AHEAD AND DO THAT.
WE'RE GOING TO DO THIS FROM THE HOME SCREEN.
SO WE'RE GOING TO PRESS 2nd QUIT.
REMEMBER POINTS OF INFLECTION
ARE POINTS ON THE ORIGINAL FUNCTION
SO THESE Y VALUES MUST COME FROM THE ORIGINAL FUNCTION,
WHICH WE PUT IN Y1.
SO WE'RE GOING TO PRESS VARS, RIGHT ARROW ONCE,
ENTER, AND THEN ENTER AND THEN IN PARENTHESES
WE CAN ENTER THE X VALUE.
SO WE'LL OPEN THIS PARENTHESIS,
2 - THEN WE'LL HAVE 2 SQUARE ROOT 3 DIVIDED BY 3.
AND THEN PRESS ENTER AND BECAUSE WE USE THE EXACT X VALUE
THIS WOULD BE THE EXACT Y VALUE
BUT WE'LL GO AHEAD AND ROUND THIS TO THREE DECIMAL PLACES
AS WE DID HERE SO -21.333.
LET'S GO AHEAD AND FIND THE Y COORDINATE
OF THE SECOND POINT OF INFLECTION.
WE CAN MAKE THIS CHANGE PRETTY EASILY.
IF WE PRESS 2nd ENTER WE CAN GO BACK AND EDIT THIS.
REMEMBER THE ONLY CHANGE WOULD BE TO CHANGE
THE SUBTRACTION HERE TO ADDITION.
PRESS ENTER AND NOTICE THAT BOTH Y COORDINATES ARE THE SAME.
NOW IF WE WOULD HAVE USED THESE ROUNDED X VALUES
OUR Y VALUES WOULD BE OFF SLIGHTLY.
SO IF YOU HAVE ONLINE HOMEWORK
SOMETIMES IT CAN BE A LITTLE TRICKY AS TO WHAT THE COMPUTER
WANTS YOU TO DO BUT SINCE WE USE
THE EXACT X VALUE THESE Y VALUES WOULD BE MORE PRECISE.
LET'S GO AND TAKE A LOOK AT A GRAPH OF THE ORIGINAL FUNCTION.
HERE ARE THE TWO RELATIVE MAXIMUMS
THAT WE FOUND AS WELL AS THE RELATIVE MINIMUM
AND YOU CAN SEE RIGHT IN HERE JUST LESS THAN ONE;
WE HAVE A POINT OF INFLECTION.
IT CHANGES FROM CONCAVE DOWN TO CONCAVE UP
WITH ANOTHER POINT OF INFLECTION ROUGHLY IN THIS AREA HERE.
SO THIS GRAPHICALLY VERIFIES OUR WORK.
I HOPE YOU FOUND THIS VIDEO HELPFUL.
THANK YOU FOR WATCHING.