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- WELCOME TO A SECOND VIDEO ON THE CONCAVITY OF A FUNCTION.
THIS VIDEO PROVIDES ADDITIONAL EXAMPLES
USING TRANSCENDENTAL FUNCTIONS.
YOU MAY WANT TO WATCH THE PREVIOUS VIDEO FIRST
BECAUSE IT DOES EXPLAIN THE CONCEPT OF CONCAVITY
IN A LITTLE MORE DETAIL.
LET'S GO AHEAD AND DO A QUICK REVIEW.
A TEST FOR CONCAVITY DEALS WITH THE SECOND DERIVATIVE.
IF THE SECOND DERIVATIVE OF THE FUNCTION IS POSITIVE
OR GREATER THAN ZERO FOR ALL X IN THE INTERVAL
THEN THE GRAPH IS CONCAVE UPWARD IN THE INTERVAL.
SO IF THE SECOND DERIVATIVE IS POSITIVE
THE FUNCTION IS CONCAVE UP.
IF THE SECOND DERIVATIVE IS LESS THAN ZERO OR NEGATIVE
FOR ALL X IN AN INTERVAL "I"
THEN THE GRAPH OF THE FUNCTION IS CONCAVE DOWNWARD
IN THE INTERVAL.
SO IF THE SECOND DERIVATIVE IS NEGATIVE
THE FUNCTION IS CONCAVE DOWN.
SO HERE'S HOW WE'RE GOING TO DETERMINE CONCAVITY.
WE'LL DETERMINE THE VALUES
FOR WHICH THE SECOND DERIVATIVE IS EQUAL TO ZERO OR UNDEFINED.
THESE WILL ALSO BE OUR POSSIBLE POINTS OF INFLECTION.
NEXT WE'LL USE THE VALUES TO DETERMINE TEST INTERVALS
BASED UPON THE DOMAIN OR THE GIVEN INTERVAL.
THEN WE'LL DETERMINE THE SIGN OF THE SECOND DERIVATIVE
IN EACH INTERVAL.
AGAIN, IF THE SECOND DERIVATIVE IS POSITIVE IT'S CONCAVE UP.
IF THE SECOND DERIVATIVE IS NEGATIVE IT'S CONCAVE DOWN.
AND IF THE INTERVALS CHANGE SIGN,
THERE IS A POINT OF INFLECTION.
LET'S GO AHEAD AND DO SOME EXAMPLES.
SO THE FIRST THING WE NEED TO DO IS FIND THE SECOND DERIVATIVE
AND DETERMINE WHERE ITS EQUAL TO ZERO OR UNDEFINED.
SO WE'LL START BY FINDING THE FIRST DERIVATIVE.
DERIVATIVE OF SINE X WOULD BE COSINE X.
DERIVATIVE OF COSINE X WOULD BE -SINE X.
THAT'S OUR FIRST DERIVATIVE.
NOW WE CAN FIND THE SECOND DERIVATIVE.
DERIVATIVE OF COSINE X WOULD BE -SINE X - DERIVATIVE OF SINE X
WHICH IS COSINE X.
AGAIN, WE WANT TO KNOW WHEN THIS IS EQUAL TO ZERO OR UNDEFINED.
IT'S NEVER UNDEFINED, SO NOW WE NEED TO SOLVE THIS EQUATION.
WE HAVE TWO DIFFERENT TRIG FUNCTIONS HERE,
SO WHAT WE CAN DO IS ADD SINE X TO BOTH SIDES.
AND LET'S GO AHEAD AND DIVIDE BY COSINE X ON BOTH SIDES.
WHAT WE'RE LEFT WITH IS -1 ON THE LEFT EQUALS--
REMEMBER SINE X DIVIDED BY COSINE X IS TANGENT X.
SO WE WANT TO KNOW WHEN TANGENT X = -1
ON THE OPEN INTERVAL FROM ZERO TO 2PI.
WELL, HAVING A TANGENT FUNCTION VALUE OF -1
SHOULD REMIND YOU HAVING A 45 DEGREE REFERENCE ANGLE.
AND SINCE THE TANGENT FUNCTION VALUE IS NEGATIVE
WE'LL SKETCH A 45 DEGREE REFERENCE ANGLE
IN THE SECOND AND THE FOURTH QUADRANT.
THEREFORE, SO THE SECOND DERIVATIVE WOULD BE ZERO
WHEN X = 3PI/4 RADIANS,
AND ALSO WHEN X = 7PI/4 RADIANS.
SO NOW WHAT WE'LL DO IS TAKE THE INTERVAL FROM ZERO TO 2PI
AND BREAK IT UP INTO THREE INTERVALS
USING THESE TWO X VALUES.
SO IT WOULD LOOK SOMETHING LIKE THIS.
OUR FIRST INTERVAL WOULD BE FROM ZERO TO 3PI/4,
THE SECOND INTERVAL WOULD BE FROM 3PI/4 TO 7PI/4,
AND THE THIRD INTERVAL WOULD BE FROM 7PI/4 TO 2PI.
REMEMBER THE SECOND DERIVATIVE WAS EQUAL TO ZERO
WHEN X = 3PI/4 RIGHT HERE, AND ALSO WHEN X = 7PI/4.
OKAY, NOW WE NEED TO DETERMINE
THE SIGN OF THE SECOND DERIVATIVE
IN EACH OF THESE INTERVALS.
LET'S GO AHEAD AND DO THAT ON THE GRAPHING CALCULATOR. OKAY.
SO WE'LL TYPE THE SECOND DERIVATIVE INTO Y1.
AND NOW WE'LL USE THE TABLE FEATURE
TO DETERMINE WHETHER THIS WOULD BE POSITIVE OR NEGATIVE
IN EACH OF THESE INTERVALS.
SO WE'LL PRESS SECOND, GRAPH,
AND NOW WE'LL TYPE IN OUR TEST VALUES.
WE NEED A VALUE BETWEEN ZERO AND 3PI/4.
HOW ABOUT PI/2?
NOTICE IT'S NEGATIVE.
NOW WE NEED A VALUE BETWEEN 3PI/4 AND 7PI/4.
HOW ABOUT PI?
NOTICE IT'S POSITIVE.
AND THE LAST INTERVAL WAS FROM 7PI/4 TO 2PI.
I KNOW 2PI IS APPROXIMATELY 6.28,
SO LET'S USE A TEST VALUE OF 6,
AND YOU CAN SEE THAT IT'S NEGATIVE.
SO WE CAN SEE THE INTERVALS ARE NEGATIVE, POSITIVE,
AND THEN NEGATIVE.
LET'S GO AHEAD AND RECORD THAT.
SO WHAT THAT TELLS US IS ON THIS INTERVAL
SINCE THE SECOND DERIVATIVE IS NEGATIVE
THE FUNCTION IS CONCAVE DOWN.
AND IT'S CONCAVE UP IN THIS ENTIRE INTERVAL.
AND IT'S CONCAVE DOWN IN THIS ENTIRE INTERVAL.
AND THE LAST THING WE HAVE TO DO
IS DETERMINE POINTS OF INFLECTION.
REMEMBER POINTS OF INFLECTION OCCUR
WHERE THE FUNCTION CHANGES CONCAVITY.
NOTICE IT CHANGES FROM CONCAVE DOWN TO CONCAVE UP,
THEREFORE AT X = 3PI/4 WE HAVE A POINT OF INFLECTION,
AND ALSO AT 7PI/4.
SO THESE ARE POINTS OF INFLECTION.
NOW, WE STILL HAVE TO GO BACK AND FIND THE Y COORDINATES
OF THESE POINTS OF INFLECTION.
AND BE CAREFUL, THESE ARE POINTS ON THE FUNCTION,
SO WE HAVE TO GO BACK TO THE ORIGINAL FUNCTION
TO DETERMINE THE FUNCTION VALUE.
SO WE NEED TO FIND F OF 3PI/4 AND F OF 7PI/4.
SO USING OUR REFERENCE TRIANGLE FOR 3PI/4,
THE SIGN OF 3PI/4 WOULD BE 1/SQUARE ROOT 2.
AND THE COSINE OF 3PI/4 WOULD BE -1 SQUARE ROOT 2.
SO THE Y COORDINATE WOULD BE ZERO HERE.
AND THEN FOR THE SIGN OF 7PI/4
WE'RE GOING TO HAVE -1 SQUARE ROOT 2
+ THE COSINE OF 7PI/4 + + 1/SQUARE ROOT 2
THAT'S ALSO EQUAL TO ZERO.
LET'S GO AHEAD AND VERIFY ALL THIS INFORMATION WITH A GRAPH.
HERE ARE TWO POINTS OF INFLECTION.
WE CAN SEE IT CHANGES CONCAVITY AT THESE POINTS,
AND ALSO THAT IN THIS INTERVAL HERE IT'S CONCAVE DOWN.
HERE IT'S CONCAVE UP,
AND THIS LITTLE PIECE HERE IS CONCAVE DOWN.
SO ALL OF OUR WORK LOOKS GOOD.
LET'S GO AND TRY ONE MORE EXAMPLE.
SAME QUESTION,
WE WANT TO DETERMINE WHERE THE FUNCTION IS CONCAVE UP OR DOWN,
AND ALSO STATE POINTS OF INFLECTION.
SO WE'LL START BY FINDING THE FIRST DERIVATIVE.
THIS REQUIRES THE PRODUCT RULE.
WE HAVE THE FIRST x DERIVATIVE OF THE SECOND THAT BE 1/X
+ THE SECOND X THE DERIVATIVE OF THE FIRST THAT WOULD BE 2X.
OKAY. SO THIS IS EQUAL TO X + 2X NATURAL LOG X.
NOW LET'S GO AHEAD AND FIND THE SECOND DERIVATIVE.
SO DERIVATIVE OF X WOULD BE 1 +
THIS WILL REQUIRE THE PRODUCT RULE
SO WE'LL HAVE 2X x DERIVATIVE OF NATURAL LOG X THAT'S 1/X
+ THE SECOND x DERIVATIVE OF 2X WHICH IS 2.
SO THIS X SIMPLIFIES WITH THIS X,
SO WE'D HAVE 1 + 2 THAT'S 3 + THIS WOULD BE 2 NATURAL LOG X.
WE WANT TO KNOW WHEN THIS IS EQUAL TO ZERO OR UNDEFINED.
WELL, THE ORIGINAL FUNCTION HAS A DOMAIN
WHERE X IS GREATER THAN ZERO.
SO ON THE ORIGINAL DOMAIN THIS WILL NEVER BE UNDEFINED.
SO TO SOLVE THIS FOR X, WHAT WE CAN DO IS
SUBTRACT 3 ON BOTH SIDES AND THEN DIVIDE BY 2.
THAT WILL GIVE US NATURAL LOG X = -3/2.
REMEMBER NATURAL LOG HAS BASE E,
SO IF WE REWRITE THIS IN EXPONENTIAL FORM
WE WOULD HAVE E TO THE POWER OF -3/2 = X.
AND THIS IS THE ONLY VALUE IN THE ORIGINAL DOMAIN
WHERE THE SECOND DERIVATIVE WOULD EQUAL ZERO.
SO THE INTERVALS WE'LL CONSIDER
WILL BE FROM ZERO TO E TO THE -3/2 POWER,
AND THEN FROM E TO THE -3/2 POWER TO +INFINITY.
LET'S GET A DECIMAL APPROXIMATION
FOR E TO THE -3/2 POWER.
SO THAT'S APPROXIMATELY 0.223.
LET'S GO AHEAD AND WRITE THAT DOWN.
THAT'S HELPFUL BECAUSE WE DO HAVE TO PICK A TEST VALUE
IN EACH INTERVAL.
LET'S GO AHEAD AND DETERMINE THE SIGN OF THESE INTERVALS
USING THE CALCULATOR.
SO WE'LL GO TO Y1 AND TYPE IN OUR SECOND DERIVATIVE
WHICH WAS 3 + 2 NATURAL LOG X.
LET'S GO TO OUR TABLE FEATURE.
WE NEED A VALUE BETWEEN ZERO AND APPROXIMATELY 0.223.
LET'S CHOOSE 0.1.
NOTICE IT'S NEGATIVE.
AND THEN WE NEED A VALUE IN THE SECOND INTERVAL.
LET'S GO AHEAD AND JUST USE THE VALUE 1.
SO NOTICE IT'S NEGATIVE IN THE FIRST INTERVAL
AND POSITIVE IN THE SECOND INTERVAL.
LET'S GO AHEAD AND RECORD THAT, NEGATIVE HERE AND POSITIVE HERE.
THAT MEANS IT'S CONCAVE DOWN IN THIS INTERVAL
AND IT'S CONCAVE UP IN THIS INTERVAL.
SO WE DO HAVE A POINT OF INFLECTION
WHERE THE X COORDINATE WOULD BE E TO THE -3/2 POWER.
AND WE HAVE TO FIND THE FUNCTION VALUE
USING THE ORIGINAL FUNCTION.
AGAIN, LET'S GO AHEAD AND JUST USE THE CALCULATOR FOR THAT.
HERE'S OUR ORIGINAL FUNCTION,
SO WE'RE GOING TO HAVE E RAISED TO THE POWER OF -3/2 SQUARED
x NATURAL LOG E TO THE POWER OF -3/2.
SO THE Y COORDINATE OF OUR POINT OF INFLECTION
WOULD BE APPROXIMATELY -0.075.
LET'S GO AHEAD AND VERIFY ALL THIS INFORMATION WITH A GRAPH.
IT IS A LITTLE DIFFICULT TO TELL,
BUT RIGHT AROUND HERE WE DO HAVE A POINT OF INFLECTION
RIGHT BELOW THE X-AXIS,
SO THAT DOES SEEM TO VERIFY OUR POINT OF INFLECTION
AS WE FOUND ON THE PREVIOUS SCREEN.
WHERE THIS VERY SMALL INTERVAL HERE WOULD BE CONCAVE DOWN
AND THIS LARGE INTERVAL HERE WOULD BE CONCAVE UP.
OKAY. I HOPE YOU FOUND THESE ADDITIONAL EXAMPLES HELPFUL.
THANK YOU FOR WATCHING.