- IN THIS EXAMPLE WE WANT TO DETERMINE ANY RELATIVE EXTREMA
USING THE SECOND DERIVATIVE TEST.
SO THE FIRST STEP IS DETERMINE THE CRITICAL NUMBERS
OF THE FUNCTION,
WHICH IS WHERE THE FIRST DERIVATIVE IS EQUAL TO ZERO
OR UNDEFINED.
SO F PRIME OF X IS GOING TO BE EQUAL TO 4X TO THE 3RD -
AND THE DERIVATIVE OF 4/3X CUBED WOULD BE 4X SQUARED.
THERE ARE NO VALUES OF X WHERE THIS WILL BE UNDEFINED,
SO LET'S GO AHEAD AND SET IT EQUAL TO ZERO
TO DETERMINE THE CRITICAL NUMBERS.
NOW, THIS IS FACTORABLE.
THE GREATEST COMMON FACTOR OF THESE TWO TERMS
WOULD BE 4X SQUARED LEAVING US WITH X - 1.
SO FOR THIS TO EQUAL 0,
EITHER 4X SQUARED MUST EQUAL 0 OR X - 1 MUST EQUAL 0.
SO FROM THIS EQUATION WE HAVE X = 0
AND FROM THIS EQUATION WE HAVE X = 1.
SO, AGAIN, THESE TWO VALUES ARE THE CRITICAL NUMBERS,
WHICH ARE THE POSSIBLE LOCATIONS FOR ANY RELATIVE MAXIMUMS
OR RELATIVE MINIMUMS.
SO TO USE THE SECOND DERIVATIVE TEST
WE'LL SUBSTITUTE THESE X VALUES INTO THE SECOND DERIVATIVE.
AND IF THE SECOND DERIVATIVE IS POSITIVE
THE FUNCTION WILL BE CONCAVE UP,
TELLING US WE HAVE A RELATIVE MINIMUM.
AND IF THE SECOND DERIVATIVE IS NEGATIVE
THAT TELLS US THE FUNCTION IS CONCAVE DOWN
AT THAT CRITICAL NUMBER,
THEREFORE WE HAVE A RELATIVE MAXIMUM.
SO USING THIS FORM OF THE FIRST DERIVATIVE
MUST DETERMINE THE SECOND DERIVATIVE.
F DOUBLE PRIME OF X IS GOING TO BE EQUAL TO 12X SQUARED - 8X.
SO NOW WE'LL EVALUATE THE SECOND DERIVATIVE AT 0 AND AT 1.
SO LET'S GO AHEAD AND ORGANIZE THIS IN A TABLE.
HERE ARE THE TWO CRITICAL NUMBERS
AND HERE'S OUR SECOND DERIVATIVE FROM THE PREVIOUS SLIDE.
SO WE NEED TO DETERMINE THE SIGN OF F DOUBLE PRIME OF 0
AND F DOUBLE PRIME OF 1.
NOTICE WHEN X IS EQUAL TO 0
THE SECOND DERIVATIVE IS ALSO EQUAL TO 0,
WHICH MEANS THE SECOND DERIVATIVE TEST FAILS.
SO WE DON'T KNOW IF WE HAVE A RELATIVE MAX OR MIN AT X = 0.
SO WE'LL COME BACK TO THIS IN A MOMENT.
LET'S GO AND EVALUATE F DOUBLE PRIME OF 1.
WE WOULD HAVE 12 x 1 SQUARED - 8 x 1,
IT'S GOING TO GIVE US 4.
SO THE SECOND DERIVATIVE IS POSITIVE AT X = 1.
SO, AGAIN, AT 0 THE SECOND DERIVATIVE WAS 0,
SO WE DON'T KNOW WHAT'S HAPPENING THERE.
BUT X = 1 THE SECOND DERIVATIVE WAS POSITIVE,
WHICH MEANS THE FUNCTION IS CONCAVE UP,
SO IT LOOKS LIKE THIS.
SO AT X = 1 WE WOULD HAVE A RELATIVE MINIMUM.
SO LET'S GO AHEAD AND RECORD THIS DOWN HERE.
THE RELATIVE MINIMUM, WE DON'T KNOW YET,
BUT IT IS LOCATED AT X = 1.
SO TO DETERMINE THE ACTUAL RELATIVE MINIMUM VALUE
WE DO HAVE TO SUBSTITUTE X = 1 BACK INTO THE ORIGINAL FUNCTION,
AND LET'S DO THAT NOW.
F OF 1 IS GOING TO BE EQUAL TO 1 TO THE 4TH - 4/3 x 1 TO THE 3RD.
WELL, THAT'S GOING TO BE 1 - 4/3 WHICH IS GOING TO BE -1/3.
SO THE RELATIVE MINIMUM VALUE IS -1/3 AT X = 1.
SOMETIMES THAT'S WRITTEN AS AND ORDERED PAIR,
WHICH WOULD BE THE POINT (1,- 1/3).
NOW, SINCE THE FIRST DERIVATIVE TEST DID FAIL AT X = 0
WE ARE SUPPOSED TO USE THE FIRST DERIVATIVE TEST
TO SEE IF THE FUNCTION CHANGES FROM INCREASING TO DECREASING
OR FROM DECREASING TO INCREASING AT X = 0.
BUT FOR THIS EXAMPLE
WE'RE GOING TO GO AHEAD AND LOOK AT IT GRAPHICALLY INSTEAD.
SO HERE'S THE GRAPH OF OUR FUNCTION.
OUR CRITICAL NUMBERS WERE X = 0 HERE AND X = 1 HERE.
SO WE CAN SEE WE DO HAVE A RELATIVE MINIMUM AT X = 1,
AND THIS WAS THE POINT (1,-1/3).
BUT NOTICE AT X = 0,
EVEN THOUGH THE DERIVATIVE WOULD BE EQUAL TO 0 HERE,
THE FUNCTION IS DECREASING TO THE LEFT OF 0,
AND NOTICE HOW IT'S ALSO DECREASING TO THE RIGHT OF 0.
SO THE FIRST DERIVATIVE WOULD BE NEGATIVE
BOTH TO THE LEFT AND RIGHT OF ZERO,
WHICH LET'S US KNOW WE DO NOT HAVE A RELATIVE EXTREMA
AT X = 0.
THE FUNCTION DOES CHANGE CONCAVITY AT X = 0,
SO THIS WOULD BE A POINT OF INFLECTION.
BUT WE'RE NOT ASKED TO PROVIDE THAT INFORMATION
IN THIS QUESTION.
SO IF WE GO BACK TO THE PREVIOUS SLIDE,
THERE'S NO RELATIVE MAXIMUM, BUT THERE IS A RELATIVE MINIMUM.
I HOPE THIS EXAMPLE WAS HELPFUL.