- WELCOME TO THE SECOND DERIVATIVE TEST.
THE GOAL OF THIS VIDEO WAS TO USE THE SECOND DERIVATIVE TEST
TO DETERMINE RELATIVE EXTREMA.
NOW, THIS PRESENTATION ASSUMES YOU'VE ALREADY WATCHED THE VIDEO
ENTITLED "CONCAVITY".
IF YOU HAVEN'T, YOU MIGHT WANT TO DO SO.
SO HERE'S HOW WE'RE GOING TO USE THE SECOND DERIVATIVE TEST
TO FIND RELATIVE EXTREMA.
IF F IS SOME FUNCTION WITH A CRITICAL VALUE C,
REMEMBER WE FIND THE CRITICAL VALUES BY SETTING THE DERIVATIVE
EQUAL TO ZERO
OR FINDING WHERE THE FIRST DERIVATIVE DOES NOT EXIST.
NOW, IF WE TAKE THIS CRITICAL VALUE
AND SUB IT INTO THE SECOND DERIVATIVE,
REMEMBER THE SECOND DERIVATIVE TELLS US CONCAVITY.
SO IF THE SECOND DERIVATIVE EVALUATED AT C
IS GREATER THAN ZERO OR POSITIVE,
THEN F OF C IS A RELATIVE MINIMUM.
REMEMBER IF THE SECOND DERIVATIVE IS POSITIVE
THE FUNCTION IS CONCAVE UP.
THEREFORE, OUR CRITICAL VALUE WOULD REPRESENT THE LOCATION
OF A MINIMUM,
AND A Y VALUE OF THIS POINT WOULD BE THE RELATIVE MINIMUM.
AND IF WE TAKE THE CRITICAL NUMBER
AND SUB IT INTO THE SECOND DERIVATIVE
AND IT'S LESS THAN ZERO OR NEGATIVE
THAT MEANS THAT THE FUNCTION WILL BE CONCAVE DOWN.
AND IF THE FUNCTION IS CONCAVE DOWN AT A CRITICAL VALUE
THAT MEANS THAT VALUE OF C WOULD HAVE A HIGH POINT,
SO F OF C WOULD BE A RELATIVE MAXIMUM.
SO IT'S GOING TO BE A LOT LESS WORK TO FINDING RELATIVE EXTREMA
USING THE SECOND DERIVATIVE THAN THE FIRST DERIVATIVE.
THE ONLY DRAWBACK IS THAT SOMETIMES THE TEST FAILS.
IF THE SECOND DERIVATIVE EVALUATED AT THE CRITICAL VALUES
EQUAL TO ZERO, TEST FAILS,
AND WE HAVE TO REVERT BACK TO THE FIRST DERIVATIVE TEST.
SO LET'S KIND OF SUMMARIZE THE PROCEDURE
FOR FINDING RELATIVE EXTREMA USING THE SECOND DERIVATIVE.
STEP ONE, WE'LL FIND THE CRITICAL VALUES OF THE FUNCTION
BY DETERMINING WHERE THE FIRST DERIVATIVE IS EQUAL TO ZERO
OR DOES NOT EXIST.
STEP TWO, WE'LL FIND THE CORRESPONDING Y VALUES
OF THE CRITICAL VALUES.
STEP THREE, WE'LL FIND THE SECOND DERIVATIVE
AND DETERMINE THE SINE OF THE SECOND DERIVATIVE
AT THE CRITICAL VALUES.
AND AS JUST STATED, IF THE SECOND DERIVATIVE IS POSITIVE
OR GREATER THAN ZERO,
WE HAVE A RELATIVE MINIMUM.
AND IF THE SECOND DERIVATIVE IS LESS THAN ZERO OR NEGATIVE,
THEN THERE IS A RELATIVE MAXIMUM.
AND THE TEST FAILS IF THE SECOND DERIVATIVE IS EQUAL TO ZERO.
LET'S GO AHEAD AND GIVE IT A TRY.
SO WE WANT TO USE THE SECOND DERIVATIVE TEST
TO FIND ALL RELATIVE EXTREMA.
SO WE'RE GOING TO FIRST FIND THE CRITICAL VALUES
BY FINDING THE FIRST DERIVATIVE.
THERE ARE NO VALUES OR THIS FIRST DERIVATIVE DOES NOT EXIST.
BUT THERE ARE VALUES WHERE IT'S GOING TO EQUAL ZERO,
THIS FACTORABLE.
SO OUR CRITICAL VALUES ARE X = 0 AND X = 2.
THESE WOULD BE THE X COORDINATES OF THE TWO POINTS
WE'RE GOING TO CONSIDER FOR A RELATIVE MAX OR MIN.
NEXT WE NEED TO FIND THE CORRESPONDING Y VALUES
TO THESE POINTS ON THE FUNCTION.
LET'S GO AHEAD AND USE THE GRAPHING CALCULATOR TO DO THIS.
YOU CAN SEE HERE I'VE ALREADY TYPED THE ORIGINAL FUNCTION
INTO Y1.
SO LET'S USE THE TABLE FEATURE TO FIND THE Y VALUES AT X = 0
AND X = 2.
SO IT LOOKS LIKE ONE POINT IS (0,-4)
AND THE OTHER ONE IS (2,0).
NEXT WE WANT TO FIND THE SECOND DERIVATIVE
AND THEN DETERMINE THE SIGN AT THESE CRITICAL VALUES.
SO LET'S GO AHEAD AND FIND THE SECOND DERIVATIVE.
WELL, HERE'S THE FIRST DERIVATIVE,
SO THE SECOND DERIVATIVE WOULD BE EQUAL TO -6X + 6.
SO WE'LL EVALUATE THESE BY HAND, IT'S PRETTY STRAIGHT FORWARD.
SO HERE WE WOULD HAVE -6 x 0 + 6.
THIS WILL BE A POSITIVE VALUE,
SO LET'S RECORD THE SIGN OF THE SECOND DERIVATIVE.
NEXT WE NEED TO EVALUATE THE SECOND DERIVATIVE AT X = 2.
AGAIN, WE DON'T CARE ABOUT THE SPECIFIC VALUE.
WE JUST CARE ABOUT THE SIGN. THIS WOULD BE -12 + 6.
THE RESULT WOULD BE NEGATIVE,
SO OUR CONCLUSION HERE IS THE FUNCTION WOULD BE CONCAVE UP
AT THIS POINT.
SO IF ITS CONCAVE UP
WE CAN CONCLUDE THAT WE WOULD HAVE A RELATIVE MINIMUM.
SO LET'S GO AHEAD AND RECORD THAT.
THE RELATIVE MINIMUM WILL BE THE Y VALUE.
SO THE RELATIVE MINIMUM IS EQUAL TO -4 AT X = 0.
ON THIS INTERVAL THE SECOND DERIVATIVE IS NEGATIVE,
SO IT'S CONCAVE DOWN.
SO IF IT'S CONCAVE DOWN IT WOULD LOOK LIKE THIS.
SO WE SHOULD BE ABLE TO TELL THAT THAT MEANS
WE'RE GOING TO HAVE A RELATIVE MAXIMUM.
SO WE WOULD STATE THAT OUR RELATIVE MAXIMUM IS EQUAL TO 0,
THE Y VALUE, WHEN X IS EQUAL TO 2.
SO AS YOU CAN SEE, IT'S A LOT LESS WORK
THAN USING THE FIRST DERIVATIVE TEST
TO DETERMINE RELATIVE EXTREMA.
LET'S GO AHEAD AND VERIFY THIS WITH A GRAPH.
HERE'S OUR RELATIVE MINIMUM OR OUR LOW POINT THAT WE FOUND.
HERE'S OUR RELATIVE MAXIMUM OR THE HIGH POINT THAT WE FOUND.
SO OUR WORK LOOKS GOOD.
LET'S GO AHEAD AND TRY ONE MORE EXAMPLE.
THIS ONE'S A LITTLE MORE INVOLVED.
STEP ONE, OF COURSE WE'RE GOING TO FIND THE FIRST DERIVATIVE
AND SET IT EQUAL TO ZERO.
SO WE HAVE -15X TO THE 4TH + 15X SQUARED.
AGAIN, THIS EXIST OVER ALL REALS,
SO WE KNOW WE'RE GOING TO FIND OUR CRITICAL VALUES
AND SET IT EQUAL TO ZERO AND FACTOR.
SO DIFFERENCE OF SQUARES.
SO WE HAVE THREE CRITICAL VALUES OFTEN CALLED CRITICAL NUMBERS.
WE HAVE X = 0, X = -1, X = +1.
THESE WILL BE THE X COORDINATES OF THE THREE POINTS
WE'LL CONSIDER FOR A RELATIVE MAX AND MIN.
SO TO FIND THE Y COORDINATES OF THESE POINTS
WE'LL USE THE GRAPHING CALCULATOR.
WE'LL TYPE THE ORIGINAL FUNCTION INTO THE GRAPHING CALCULATOR
AND USE THE TABLE FEATURE AGAIN.
NOW, I'VE ALREADY TYPED IT INTO Y2,
SO WHEN WE GO TO THE TABLE WE'LL JUST LOOK AT THE Y2 COLUMN.
AGAIN, OUR X VALUES ARE 0, -1, AND 1.
SO LOOKS LIKE WE HAVE (0,0), (-1,-2), AND (1,2).
NEXT WE NEED TO FIND THE SECOND DERIVATIVE
SO WE CAN DETERMINE THE SIGN OF THE SECOND DERIVATIVE
AT THESE POINTS.
SO IF HERE'S OUR FIRST DERIVATIVE,
OUR SECOND DERIVATIVE WOULD EQUAL -60X CUBED + 30X.
SO WE NEED TO DETERMINE THE SIGN THE SECOND DERIVATIVE
AT X = 0, AT X = -1, AND AT X = +1.
SO LET'S GO BACK TO THE GRAPHING CALCULATOR
AND TYPE IN OUR SECOND DERIVATIVE--
AND TYPE IN OUR SECOND DERIVATIVE
AND GO BACK TO THE TABLE.
SO AT 0, SECOND DERIVATIVE IS EQUAL TO 0.
REMEMBER THAT TELLS US THE TEST FAILS.
AT -1 THE SECOND DERIVATIVE IS POSITIVE,
AND AT + 1 IT'S NEGATIVE.
SO LET'S RECORD THIS INFORMATION.
X = 1, IT WAS POSITIVE, AT X = +1, IT WAS NEGATIVE.
SO HERE WE RECORD THAT INFORMATION TO TABLE.
LET'S DRAW OUR CONCLUSION.
HERE THE TEST FAILS, HERE IT'S CONCAVE UP.
SO IF IT'S CONCAVE UP YOU WOULD HAVE A RELATIVE MINIMUM OF -2.
LET'S GO AHEAD AND WRITE THAT IN.
RELATIVE MINIMUM OF -2 AT X = - 1.
AND ON THIS LAST ONE IT'S NEGATIVE--
ON THIS LAST ONE THE SECOND DERIVATIVE IS NEGATIVE,
SO IT'S CONCAVE DOWNWARD.
SO IF IT'S CONCAVE DOWNWARD IT WOULD LOOK LIKE THIS.
SO WE CAN TELL WE'D HAVE A HIGH POINT OR RELATIVE MAXIMUM.
AND SO OUR RELATIVE MAX WOULD BE EQUAL TO 2 AT X = 1.
NOW, WE ARE SUPPOSED TO GO BACK AND USE THE FIRST DERIVATIVE
TO ANALYZE THIS CASE MORE CAREFULLY,
BUT LET'S GO AHEAD AND JUST TAKE A LOOK AT THE GRAPH
AND SEE WHAT WE CAN DETERMINE.
SO HERE'S OUR RELATIVE MINIMUM THAT WE FOUND.
HERE'S OUR RELATIVE MAXIMUM.
HERE'S THE POINT THAT FAILED THE TEST
AND YOU CAN SEE THIS IS NOT A RELATIVE MAX OR RELATIVE MIN.
TO THE LEFT OF THIS POINT THE FUNCTION IS INCREASING
AND IT'S ALSO INCREASING TO THE RIGHT.
SO, IN FACT, WE DO NOT HAVE A RELATIVE EXTREMA
AT THE POINT (0,0).
I HOPE THAT HELPS EXPLAIN
HOW YOU CAN USE A SECOND DERIVATIVE TEST
TO FIND RELATIVE EXTREMA.
THANK YOU FOR WATCHING.