- WE WANT TO FIND 
THE RELATIVE EXTREMA
OR RELATIVE MAX AND MIN'S 
OF F OF X = X + 2 DIVIDED BY X.
SO FOR THE FIRST STEP WE WANT TO 
FIND THE CRITICAL NUMBERS
WHICH WOULD BE THE LOCATIONS OF 
THE POSSIBLE RELATIVE EXTREMA.
THE CRITICAL NUMBERS OCCUR
WHERE THE FIRST DERIVATIVE 
IS EQUAL TO ZERO OR UNDEFINED.
BEFORE WE FIND THE DERIVATIVE 
FUNCTION
LET'S REWRITE THE GIVEN FUNCTION 
AS F OF X = X + 2
x X TO THE POWER OF -1
SO THAT WE CAN APPLY 
THE POWER RULE
TO FIND OUR DERIVATIVE FUNCTION.
REMEMBER THIS IS X TO THE FIRST
SO IF WE MOVE THIS 
ACROSS THE FRACTION BAR
IT'S GOING TO CHANGE THE SIGN 
OF THE EXPONENT.
SO F PRIME OF X IS EQUAL TO THE 
DERIVATIVE OF X WHICH IS 1
+ THE DERIVATIVE OF 2X 
TO THE -1.
SO WE'RE GOING TO MULTIPLY 
BY THE EXPONENT.
THAT'S GOING TO BE -2 OR - 2 X 
TO THE POWER OF -1 - 1 IS -2.
SO WE WANT TO KNOW 
WHEN THIS DERIVATIVE FUNCTION
IS EQUAL TO ZERO OR UNDEFINED.
SO WE HAVE 1 - 2 
DIVIDED BY X SQUARED = 0.
IN THIS FORM WE CAN SEE THE 
DERIVATIVE WOULD BE UNDEFINED
WHEN X = 0
BUT SINCE X IS NOT IN THE DOMAIN 
OF THE ORIGINAL FUNCTION
ZERO WILL NOT BE A CRITICAL 
NUMBER FOR F OF X.
SO WE'LL GO AHEAD AND SOLVE THIS 
EQUATION HERE.
SO WE'D HAVE 1 = 2 
DIVIDED BY X SQUARED.
LET'S GO AHEAD AND MULTIPLY 
BOTH SIDES BY X SQUARED.
THIS WOULD GIVE US THE EQUATION 
X SQUARED = 2.
SO NOW WE'LL SQUARE ROOT 
BOTH SIDES OF THE EQUATION
AND SINCE X CAN BE BOTH POSITIVE 
OR NEGATIVE
WE HAVE X 
= + OR - SQUARE ROOT 2.
THE SQUARE ROOT OF X SQUARED
IS ACTUALLY EQUAL 
TO THE ABSOLUTE VALUE OF X
AND THAT'S THE REASON WHY X CAN 
BE BOTH POSITIVE OR NEGATIVE.
SO WE HAVE TWO CRITICAL NUMBERS.
ONE IS X = -SQUARE ROOT 2
AND THE SECOND CRITICAL NUMBER 
IS X = SQUARE ROOT 2.
NOW TO DETERMINE WHETHER WE HAVE 
RELATIVE MAX OR MIN'S
AT THESE LOCATIONS
WE CAN EITHER USE 
THE FIRST DERIVATIVE
OR THE SECOND DERIVATIVE TEST.
FOR THIS EXAMPLE, 
LET'S GO AHEAD AND USE
THE SECOND DERIVATIVE TEST.
SO TO FIND THE SECOND DERIVATIVE
WE'LL FIND THE DERIVATIVE 
OF THE FIRST DERIVATIVE FUNCTION
GIVEN HERE.
SO F DOUBLE PRIME OF X 
IS EQUAL TO
THE DERIVATIVE OF 1 
WHICH IS ZERO
AND THEN + THE DERIVATIVE OF -2 
X TO THE -2.
SO WE'RE GOING TO MULTIPLY 
BY THE EXPONENT.
-2 x -2 IS 4 AND THEN SUBTRACT 1 
FROM THE EXPONENT -2 - 1 = -3.
SO OUR SECOND DERIVATIVE 
IS EQUAL TO 4
DIVIDED BY X TO THE 3rd
AND NOW WE'RE GOING TO DETERMINE 
THE SIGN
OF THE SECOND DERIVATIVE 
AT THESE CRITICAL NUMBERS
AND IF THE SECOND DERIVATIVE 
IS POSITIVE
THE FUNCTION IS CONCAVE UP
AND IF IT'S NEGATIVE 
THE FUNCTION IS CONCAVE DOWN.
WELL NOTICE WHEN X IS EQUAL TO 
-SQUARE ROOT 2
IF WE CUBE A NEGATIVE
THE VALUE WILL STILL BE NEGATIVE
AND THEREFORE AT X 
= -SQUARE ROOT 2
THE SECOND DERIVATIVE 
IS LESS THAN ZERO OR NEGATIVE
WHICH MEANS THE FUNCTION 
MUST BE CONCAVE DOWN
AT X = -SQUARE ROOT 2.
WELL IF THE FUNCTION 
IS CONCAVE DOWN
IT WOULD LOOK SOMETHING LIKE 
THIS
WHICH MEANS AT X 
= -SQUARE ROOT 2
WE'D HAVE A HIGH POINT 
OR RELATIVE MAXIMUM
AND NOTICE WHEN X 
= +SQUARE ROOT 2
IF WE CUBE A POSITIVE NUMBER
A POSITIVE DIVIDED BY A POSITIVE 
IS STILL POSITIVE.
SO IF THE SECOND DERIVATIVE IS 
POSITIVE AT X = SQUARE ROOT 2
THE FUNCTION MUST BE CONCAVE UP.
SO IF IT'S CONCAVE UP IT WOULD 
LOOK SOMETHING LIKE THIS
WHICH MEANS AT X = SQUARE ROOT 2 
WE HAVE A LOW POINT
OR A RELATIVE MINIMUM
AND NOW TO DETERMINE 
THE RELATIVE MAX/MIN VALUES
WE'RE GOING TO EVALUATE 
THE ORIGINAL FUNCTION
AT X = -SQUARE ROOT 2 
AND X = +SQUARE ROOT 2.
SO THE RELATIVE MAXIMUM VALUE
WILL BE EQUAL TO 
F OF -SQUARE ROOT 2
WHICH IS EQUAL TO -SQUARE ROOT 2 
+ 2 DIVIDED BY -SQUARE ROOT 2.
LET'S GO AHEAD AND RATIONALIZE 
THIS DENOMINATOR.
SO WE HAVE -SQUARE ROOT 2,
THEN WE'D HAVE PLUS A NEGATIVE 
OR A MINUS 2 SQUARE ROOT 2/2.
NOTICE THE 2'S SIMPLIFY OUT.
SO WE HAVE -SQUARE ROOT 2 - 
SQUARE ROOT OR -2 SQUARE ROOT 2.
SO THE RELATIVE MAXIMUM VALUE 
OR THE FUNCTION VALUE
IS -2 SQUARE ROOT 2
AND NOW TO FIND THE RELATIVE 
MINIMUM
WE'LL FIND F OF SQUARE ROOT 2.
SO WE'LL HAVE SQUARE ROOT 2 + 2 
DIVIDED BY SQUARE ROOT 2.
WE'LL RATIONALIZE 
THE DENOMINATOR.
SO WE HAVE SQUARE ROOT 2
+ THIS WOULD BE 2 SQUARE ROOT 2 
OVER 2.
AGAIN, THESE 2'S SIMPLIFY OUT
SO WE HAVE SQUARE ROOT 2 
+ SQUARE ROOT 2.
SO F OF SQUARE ROOT 2 
IS EQUAL TO 2 SQUARE ROOT 2
WHICH IS THAT RELATIVE MINIMUM 
FUNCTION VALUE.
NOW SOMETIMES WE DO EXPRESS 
THE RELATIVE EXTREMA
AS ORDERED PAIRS.
SO TO EXPRESS THIS INFORMATION 
AS A POINT
WE WOULD SAY THAT THE RELATIVE 
MAX
WOULD HAVE AN X COORDINATE 
OF -SQUARE ROOT 2
AND A Y COORDINATE OF 
-2 SQUARE ROOT 2.
BUT JUST KEEP IN MIND 
THE X COORDINATE IS A LOCATION
AND THE Y COORDINATE IS THE 
ACTUAL RELATIVE MAXIMUM VALUE
AND THEN FOR THE RELATIVE 
MINIMUM
WE'D HAVE AN X COORDINATE 
OF SQUARE ROOT 2
AND A Y COORDINATE 
OF 2 SQUARE ROOT 2.
AGAIN, THE X VALUE 
IS THE LOCATION
AND THE Y VALUE 
OR FUNCTION VALUE
IS THE RELATIVE MINIMUM VALUE.
NOW THIS MAY LOOK 
A LITTLE STRANGE
BECAUSE NOTICE HOW THE RELATIVE 
MAXIMUM VALUE IS NEGATIVE
AND THE RELATIVE MINIMUM VALUE 
IS POSITIVE.
SO LET'S GO AHEAD AND VERIFY 
THIS BY LOOKING AT THE GRAPH
OF THE FUNCTION.
NOTICE THIS POINT HERE 
IS A RELATIVE MINIMUM
BECAUSE IT'S A LOW POINT 
ON THE GRAPH
AND THIS POINT HERE 
IS A HIGH POINT ON THE GRAPH
AND THEREFORE WE HAVE 
A RELATIVE MAXIMUM.
THE COORDINATES 
OF THIS LOW POINT HERE
(SQUARE ROOT 2, 2 SQUARE ROOT 2)
AND THE COORDINATES 
OF THIS HIGH POINT
ARE (-SQUARE ROOT, 
-2 SQUARE ROOT 2)
WHICH AGAIN REPRESENTS 
OUR RELATIVE MAXIMUM.
I HOPE YOU FOUND THIS EXAMPLE 
HELPFUL.
