- 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.
