- 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.