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- WE'RE GIVEN THE FUNCTION F OF X = X x THE SQUARE ROOT
OF THE QUANTITY X SQUARED + 4 ON THE CLOSED INTERVAL
FROM -4 TO 7.
WE FIRST WANT TO DETERMINE THE OPEN INTERVALS
WHERE THE FUNCTION IS CONCAVE UP OR CONCAVE DOWN.
THEN WE'LL DETERMINE ANY POINTS OF INFLECTION
AND DETERMINE THE ABSOLUTE EXTREMA.
SO TO DETERMINE WHERE A FUNCTION IS CONCAVE UP OR CONCAVE DOWN,
WE'LL FIRST DETERMINE WHERE THE SECOND DERIVATIVE
IS EQUAL TO ZERO OR UNDEFINED.
THEN BASED UPON THOSE VALUES WE'LL DIVIDE THE GIVEN INTERVAL
INTO SUB INTERVALS AND TEST THE SIGN OF THE SECOND DERIVATIVE
IN EACH SUB INTERVAL.
WELL, IF THE SECOND DERIVATIVE IS POSITIVE
THE FUNCTION IS CONCAVE UP,
AND WHERE THE SECOND DERIVATIVE IS NEGATIVE
THE FUNCTION IS CONCAVE DOWN.
SO TO FIND THE SECOND DERIVATIVE
WE'LL BEGIN BY FINDING THE FIRST DERIVATIVE.
BUT BEFORE WE BEGIN, WE WANT TO REWRITE THE GIVEN FUNCTION
USING RATIONAL EXPONENTS.
SO THE GIVEN FUNCTION F OF X
IS EQUAL TO X x THE QUANTITY X SQUARED + 4 TO THE 1/2 POWER.
TAKEN THE SQUARE ROOT OF A QUANTITY IS EQUIVALENT
TO RAISING IT TO THE 1/2 POWER.
NOTICE IN THIS FORM WE'LL HAVE TO APPLY THE PRODUCT RULE
IN ORDER TO FIND THE FIRST DERIVATIVE.
SO F PRIME OF X IS GOING TO BE EQUAL TO THE FIRST FUNCTION X
x THE DERIVATIVE OF THE SECOND FUNCTION,
WHICH WOULD BE THE DERIVATIVE OF THE QUANTITY X SQUARED + 4
TO THE 1/2 POWER + THE SECOND FUNCTION, WHICH IS X SQUARED
+ 4 TO THE 1/2 x THE DERIVATIVE OF THE FIRST FUNCTION
WHICH IS JUST X.
SO NOW WE'LL FIND THE DERIVATIVE HERE AND THE DERIVATIVE HERE.
SO F PRIME OF X IS EQUAL TO X x--
TO FIND THIS DERIVATIVE WE'LL HAVE TO APPLY
THE EXTENDED POWER RULE,
WHICH INCLUDES THE CHAIN RULE WHERE THE INNER FUNCTION U
WOULD BE EQUAL TO X SQUARED + 4.
SO WE'D HAVE 1/2 x THE QUANTITY X SQUARED + 4 TO THE 1/2 - 1
OR -1/2 x THE DERIVATIVE OF X SQUARED + 4 WHICH WOULD BE 2X.
AND THEN WE HAVE + JUST THE QUANTITY X SQUARED + 4
TO THE 1/2 BECAUSE THE DERIVATIVE OF X WOULD BE 1.
LET'S GO AHEAD AND SIMPLIFY THIS PRODUCT HERE.
NOTICE THIS 2 AND THIS 2 SIMPLIFY TO 1.
AND BECAUSE THIS IS A NEGATIVE EXPONENT
WE'LL MOVE IT TO THE DENOMINATOR.
SO THIS WOULD BE EQUAL TO X SQUARED DIVIDED BY THE QUANTITY
X SQUARED + 4 TO THE 1/2.
AND THEN, PLUS, LET'S WRITE THIS AS THE QUANTITY X SQUARED + 4
TO THE 1/2/1.
SO IF WE WANT TO ADD THESE TWO FRACTIONS
WE'D HAVE TO OBTAIN A COMMON DENOMINATOR
WHICH WOULD BE THE QUANTITY X SQUARED + 4 TO THE 1/2.
SO LET'S GO AHEAD AND MULTIPLY THIS FRACTION HERE BY X SQUARED
+ 4 TO THE 1/2/X SQUARED + 4 TO THE 1/2.
NOTICE WHEN WE DO THIS, WE HAVE OUR COMMON DENOMINATOR
WHICH IS THE QUANTITY X SQUARED + 4 TO THE 1/2.
LOOKING AT THE NUMERATOR,
NOW WE HAVE X SQUARED + HERE WHEN WE MULTIPLY
WE'D ADD THE EXPONENTS, 1/2 + 1/2 IS 1.
SO THIS PRODUCT HERE WOULD JUST BE THE QUANTITY X SQUARED + 4.
LET'S CONTINUE ON THE NEXT SLIDE.
COMBINING LIKE TERMS IN THE NUMERATOR
WE WOULD HAVE 2X SQUARED + 4 DIVIDED BY X SQUARED + 4
TO THE 1/2.
SO TO FIND THE SECOND DERIVATIVE
WE'LL HAVE TO FIND THE DERIVATIVE
OF THE FIRST DERIVATIVE.
AND NOTICE HERE WE'LL HAVE TO APPLY THE QUOTIENT RULE
GIVEN HERE BELOW.
SO F DOUBLE PRIME OF X WOULD BE EQUAL TO--
BEGINNING WITH THE DENOMINATOR
WE'D HAVE THE DENOMINATOR SQUARED,
SO WE'D HAVE THE QUANTITY X SQUARE + 4 TO THE 1/2 SQUARED.
NOTICE HOW THIS WILL SIMPLIFY NICELY.
THEN OUR NUMERATOR IS GOING TO BE
THE DENOMINATOR x THE DERIVATIVE OF THE NUMERATOR,
WHICH WOULD BE THE DERIVATIVE OF 2X SQUARED + 4 - THE NUMERATOR
x THE DERIVATIVE OF THE DENOMINATOR.
LET'S MOVE THIS DENOMINATOR TOWARDS THE MIDDLE.
SO THE SECOND DERIVATIVE WOULD BE EQUAL TO--
THE DENOMINATOR'S JUST GOING TO BE X SQUARED + 4 TO THE 1ST.
SO WE'LL HAVE X SQUARED + 4 TO THE 1/2 x THE DERIVATIVE
OF 2X SQUARED + 4 THAT'S GOING TO BE 4X.
AND THEN MINUS THE QUANTITY 2X SQUARED + 4 x THE DERIVATIVE
OF THE QUANTITY X SQUARED + 4 TO THE 1/2 IS GOING TO BE 1/2
x THE QUANTITY X SQUARED + 4 TO THE -1/2 x THE DERIVATIVE
OF X SQUARED + 4 THAT'S 2X.
SO SIMPLIFYING HERE, THE 2 AND THE 2 SIMPLIFY OUT.
SO OUR SECOND DERIVATIVE, LETS FOCUS ON THE NUMERATOR.
HERE WE'D HAVE 4X x THE QUANTITY X SQUARED + 4 TO THE 1/2.
AND THEN HERE WE'RE GOING TO HAVE MINUS--
THE NUMERATOR'S GOING TO BE X x 2X SQUARED + 4.
HERE WE HAVE A NEGATIVE EXPONENT,
SO WE'LL MOVE THIS DOWN TO THE DENOMINATOR,
WHICH WOULD BE X SQUARED + 4 TO THE 1/2.
NOW, TO SUBTRACT THESE FRACTIONS IN THE NUMERATOR WE'D HAVE TO,
AGAIN, OBTAIN A COMMON DENOMINATOR,
WHICH WOULD BE THE QUANTITY X + 4 TO THE 1/2.
SO NOW IN THIS CASE WE'LL MULTIPLY THIS FIRST FRACTION
BY X SQUARED + 4 TO THE 1/2/X SQUARED + 4 TO THE 1/2.
LET'S GO AHEAD AND FINISH THIS ON THE NEXT SLIDE.
NOW, LOOKING AT THE NUMERATOR,
NOTICE HOW WE DO HAVE A COMMON DENOMINATOR OF THE QUANTITY
X SQUARED + 4 TO THE 1/2.
AND THEN THE NUMERATOR, NOTICE HOW HERE WE HAVE THE SAME BASE,
SO WE ADD THE EXPONENTS.
SO WE'D HAVE 4X x THE QUANTITY X SQUARED + 4 TO THE 1ST
OR JUST 4X x THE QUANTITY X SQUARED + 4,
AND THEN - X x THE QUANTITY 2X SQUARED + 4.
NOW, FOR THE NEXT STEP, INSTEAD OF DIVIDING BY THE QUANTITY
X SQUARED + 4 WE'RE GOING TO MULTIPLY BY THE RECIPROCAL
AND WE'LL ALSO SIMPLIFY THIS NUMERATOR.
NOTICE WE'D HAVE 4X CUBED + 16X AND THEN - 2X CUBED - 4X.
SO WE CAN WRITE THIS AS F DOUBLE PRIME OF X IS EQUAL TO--
AGAIN, THE FRACTION UP ON TOP WOULD BE THE DENOMINATOR
OF X SQUARED + 4 TO THE 1/2.
THE NUMERATOR'S GOING TO BE--
HERE WE HAVE 4X CUBED - 2X CUBED THAT'S 2X CUBED.
AND THEN WE HAVE + 16X - 4X THAT'S 12X.
AND NOW INSTEAD OF DIVIDING BY THE QUANTITY X SQUARED + 4,
WE'LL MULTIPLY BY THE RECIPROCAL OF 1/THE QUANTITY X SQUARED + 4.
AGAIN, THIS WOULD BE TO THE FIRST POWER.
SO FINALLY, OUR SECOND DERIVATIVE FUNCTION
IS EQUAL TO 2X TO THE 3RD + 12X
DIVIDED BY THE QUANTITY X SQUARED + 4 TO THE 1/2 + 1
OR 3/2 POWER.
SO NOW, TO DETERMINE THE CONCAVITY,
WE WANT TO BEGIN BY DETERMINING WHERE THE SECOND DERIVATIVE
IS EQUAL TO ZERO OR UNDEFINED.
NOTICE BECAUSE WE HAVE A BASE
OF X SQUARED + 4 IN THE DENOMINATOR
THIS WOULD NEVER BE ZERO,
SO WE'LL NEVER HAVE DIVISION BY ZERO.
SO THIS FRACTION IS GOING TO BE ZERO WHEN THE NUMERATOR'S
EQUAL TO ZERO.
SO IF WE SET THE NUMERATOR EQUAL TO ZERO,
NOTICE HOW WE WOULD HAVE 2X TO THE 3RD + 12X = 0,
WHICH WOULD FACTOR.
WE CAN FACTOR OUT 2X LEAVING US
WITH THE QUANTITY X SQUARED + 6 = 0.
SO EITHER 2X = 0 OR X SQUARED + 6 = 0.
WELL, X SQUARED + 6 = 0 WILL NOT GIVE US ANY REAL SOLUTIONS,
BUT 2X = 0, IF WE DIVIDE BOTH SIDES BY 2, NOTICE HOW X = 0.
SO THIS IS THE ONLY VALUE WHERE THE SECOND DERIVATIVE
IS GOING TO BE EQUAL TO 0,
SO NOW WE'LL DIVIDE THE INTERVAL FROM -4 TO 7
USING THE VALUE OF X = 0.
AND THEN WE'LL TEST THE SIGN OF THE SECOND DERIVATIVE
IN EACH OF THOSE INTERVALS.
LET'S DO THIS ON THE NEXT SLIDE.
SO BECAUSE THE SECOND DERIVATIVE WAS EQUAL TO ZERO WHEN X = 0,
WE'LL DIVIDE THE INTERVAL FROM -4 TO 7
INTO THE OPEN INTERVAL FROM -4 TO 0,
AND THE OPEN INTERVAL FROM 0 TO 7.
AND NOW WE'LL TEST THE SIGN OF THE SECOND DERIVATIVE
IN EACH OF THESE INTERVALS.
SO LET'S GO AHEAD AND USE THE TEST VALUE OF X = -1
IN THIS FIRST INTERVAL.
F DOUBLE PRIME OF -1 WOULD BE EQUAL TO--THIS WOULD BE -2 + -12
THAT'S -14 DIVIDED BY THE QUANTITY -1 SQUARED + 4
TO THE 3/2 POWER.
NOTICE HOW THIS QUOTIENT WOULD BE NEGATIVE OR LESS THAN ZERO,
WHICH MEANS THE SECOND DERIVATIVE IS NEGATIVE
OVER THIS ENTIRE INTERVAL,
AND THEREFORE THE FUNCTION IS CONCAVE DOWN
ON THIS OPEN INTERVAL.
AND NOW FOR THE SECOND INTERVAL, LET'S TEST X = 1.
SO FOR F DOUBLE PRIME OF 1,
NOTICE HOW WE'D HAVE 2 + 12 THAT'S 14
DIVIDED BY THE QUANTITY 1 SQUARED + 4 TO THE 3/2 POWER.
THIS QUOTIENT WOULD POSITIVE OR GREATER THAN ZERO,
AND THEREFORE THE SECOND DERIVATIVE IS POSITIVE
OVER THIS ENTIRE INTERVAL,
AND THEREFORE THE FUNCTION IS CONCAVE UP
ON THIS ENTIRE INTERVAL.
AND NOTICE HOW THE FUNCTION CHANGES CONCAVITY AT X = 0,
WHICH MEANS YOU HAVE A POINT OF INFLECTION AT X = 0.
AGAIN, THE FUNCTION CHANGES FROM CONCAVE DOWN TO CONCAVE UP
AT X = 0,
SO FOR THE NEXT QUESTION WHEN WE'RE ASKED TO FIND THE POINT
OF INFLECTION WE KNOW THE X COORDINATE HAS TO BE ZERO.
SO TO FIND THE Y COORDINATE OR THE POINT OF INFLECTION,
WE SUBSTITUTE X = 0 BACK INTO THE ORIGINAL FUNCTION.
REMEMBER THE POINT OF INFLECTION IS A POINT ON THE FUNCTION,
SO WE'LL HAVE TO FIND F OF 0 TO FIND THE Y COORDINATE.
SO THAT WOULD GIVE US 0 x THE SQUARE ROOT OF 0 SQUARED + 4.
WELL, THAT'S GOING TO BE 0 x THE SQUARE ROOT OF 4 OR 0 x 2,
WHICH OF COURSE IS 0.
SO THE POINT OF INFLECTION IS AT THE ORIGIN OR THE POINT (0,0).
AND THEN FINALLY, THE LAST QUESTION
WAS TO FIND THE ABSOLUTE EXTREMA, WHICH MEANS
FIND THE ABSOLUTE MAXIMUM AND ABSOLUTE MINIMUM.
TO DO THIS, WE WANT TO EVALUATE THE FUNCTION AT THE ENDPOINTS
AND ANY CRITICAL NUMBERS WHICH OCCUR WHERE THE FIRST DERIVATIVE
IS EQUAL TO ZERO OR UNDEFINED.
IF WE GO BACK AND TAKE A LOOK AT OUR FIRST DERIVATIVE,
HERE WE'RE NEVER GOING TO HAVE DIVISION BY ZERO,
SO WILL NEVER BE UNDEFINED BECAUSE X SQUARED + 4
WILL NEVER BE ZERO.
AND ALSO, X SQUARED + 4 THE NUMERATOR WILL NEVER BE ZERO,
AND THEREFORE THIS FIRST DERIVATIVE IS NEVER ZERO
OR UNDEFINED BECAUSE THERE ARE NO CRITICAL NUMBERS.
SO TO FIND THE ABSOLUTE EXTREMA WE'LL EVALUATE THE FUNCTION
AT JUST THE ENDPOINTS.
TO SAVE SOME TIME, I PAUSED THE VIDEO
AND DID SOME WORK HERE.
NOTICE HERE I FOUND F OF -4,
WHICH WAS EQUAL TO -8 SQUARE ROOT 5,
AND F OF 7 WAS EQUAL TO 7 SQUARE ROOT 53.
SO THE SMALLEST VALUE HERE,
WHICH IS -8 SQUARE ROOT 5 IS THE ABSOLUTE MINIMUM.
AND 7 SQUARE ROOT 53 WOULD BE THE ABSOLUTE MAXIMUM.
AND, THEREFORE, WE CAN SAY THE ABSOLUTE MINIMUM
IS -8 SQUARE ROOT 5, WHICH IS APPROXIMATELY -17.89,
ADD X = -4, AND THE ABSOLUTE MAXIMUM IS 7 SQUARE ROOT 53,
WHICH IS APPROXIMATELY 50.96 AT X = 7.
LET'S LOOK AT OUR RESULTS ON THE GRAPH.
FIRST, TALKING ABOUT THE CONCAVITY,
NOTICE HOW THE FUNCTION IS CONCAVE DOWN
ON THE OPEN INTERVAL FROM -4 TO 0, AS WE SEE HERE.
AND THEN THE FUNCTION IS CONCAVE UP ON THE OPEN INTERVAL
FROM 0 TO 7 HERE.
AND BECAUSE THE FUNCTION CHANGES CONCAVITY AT THE ORIGIN HERE,
NOTICE HOW THE POINT OF INFLECTION IS THE POINT (0,0).
AND THEN FINALLY, ON THE CLOSED INTERVAL,
NOTICE HOW THE SMALLEST VALUE IS APPROXIMATELY -17.89 AT -4,
AND THE MAXIMUM FUNCTION VALUE IS APPROXIMATELY 50.96 AT X = 7.
I HOPE YOU FOUND THIS HELPFUL.