- WELCOME TO A SECOND VIDEO ON THE SECOND DERIVATIVE TEST.
THIS VIDEO PROVIDES ADDITIONAL EXAMPLES
USING TRANSCENDENTAL FUNCTIONS.
LET'S GO AHEAD AND DO A QUICK REVIEW
OF THE SECOND DERIVATIVE TEST.
THE SECOND DERIVATIVE TEST IS A WAY TO FIND RELATIVE EXTREMA
USING THE SECOND DERIVATIVE INSTEAD OF THE FIRST DERIVATIVE
AND HERE'S THE IDEA.
IF WE FIND THE CRITICAL NUMBERS OF THE FUNCTION
AND THEN DETERMINE THE SIGN OF THE SECOND DERIVATIVE
AT THOSE CRITICAL NUMBERS
THAT'LL TELL US THE CONCAVITY OF THE FUNCTION.
SO IF THE SECOND DERIVATIVE IS POSITIVE AT THE CRITICAL NUMBER
THAT MEANS THE FUNCTION WILL BE CONCAVE UPWARD
THEREFORE WE'D HAVE TO HAVE A RELATIVE MIN
AT THAT CRITICAL NUMBER.
IF THE SECOND DERIVATIVE IS NEGATIVE OR LESS THAN ZERO
AT THE CRITICAL NUMBER
THAT MEANS IT'S CONCAVE DOWNWARD
THEREFORE AT THAT CRITICAL NUMBER
WE MUST HAVE A RELATIVE MAX.
SO IT CAN BE A LOT LESS WORK TO DETERMINE RELATIVE EXTREMA
BY DETERMINING THE SIGN OF THE SECOND DERIVATIVE.
IF THE SECOND DERIVATIVE EQUALS ZERO,
AT THE CRITICAL THE TEST FAILS
AND WE DO HAVE TO USE THE FIRST DERIVATIVE TEST.
SO LET'S OUTLINE OUR STEPS TO FOLLOW.
WE'LL FIRST FIND THE CRITICAL NUMBERS OF THE FUNCTION
BY DETERMINING WHERE THE FIRST DERIVATIVE IS EQUAL TO ZERO
OR DOES NOT EXIST.
THEN WE'LL FIND THE CORRESPONDING Y VALUES
OF THE CRITICAL NUMBERS.
THEN WE'LL FIND THE SECOND DERIVATIVE
AND DETERMINE THE SIGN OF IT AT THE CRITICAL NUMBERS,
AND THEN AGAIN, IF THE SECOND DERIVATIVE IS POSITIVE
THAT MEANS IT'S CONCAVE UP
SO WE HAVE A RELATIVE MINIMUM AT THE CRITICAL NUMBER,
AND IF THE SECOND DERIVATIVE IS NEGATIVE OR LESS THAN ZERO
THE FUNCTION'S CONCAVE DOWN
THEREFORE WE'D HAVE A RELATIVE MAX AT THAT CRITICAL NUMBER.
LET'S GO AHEAD AND GIVE IT A TRY.
FIRST OFF THIS FUNCTION HAS A DOMAIN FROM ZERO TO INFINITY.
OKAY. SO LET'S GO AHEAD AND FIND THE FIRST DERIVATIVE.
THAT WOULD BE 1 - 1/X.
THIS IS UNDEFINED AT ZERO BUT IT'S NOT IN THE DOMAIN.
SO WE'LL SET THE DERIVATIVE EQUAL TO ZERO
AND SOLVE THIS EQUATION.
LET'S ADD 1/X TO BOTH SIDES.
IF WE WANT WE COULD PUT THIS OVER 1.
WE HAVE A PROPORTION.
SO X x 1 WOULD BE X MUST EQUAL 1 x 1.
X = 1 IS THE ONLY CRITICAL NUMBER.
SO WHAT WE'LL DO NOW IS FIND THE Y COORDINATE OF THE POINT
ON THE FUNCTION WHERE X = 1.
SO WE NEED TO FIND F OF 1.
WELL HERE'S F THAT WOULD BE 1 - NATURAL LOG 1.
WELL NATURAL LOG 1 IS EQUAL TO ZERO.
SO THE POINT IS (1, 1).
SO NOW WHAT WE'RE GOING TO DO IS FIND THE SECOND DERIVATIVE
AND DETERMINE THE SIGN OF IT WHEN X IS EQUAL TO 1.
SO LET'S GO AHEAD AND REWRITE THE FIRST DERIVATIVE HERE.
THE FIRST DERIVATIVE WAS EQUAL TO 1 -
THIS WOULD BE X TO THE POWER OF -1.
SO THE SECOND DERIVATIVE WILL BE ZERO
AND THEN - -1 X TO THE -2 AND THAT WOULD BE +X TO THE -2
OR 1/X SQUARED.
WELL F DOUBLE PRIME OF 1, OUR CRITICAL NUMBER,
IS EQUAL TO 1/1 SQUARED WHICH IS 1.
THAT'S POSITIVE.
WHAT THAT MEANS IS THE FUNCTION IS CONCAVE UP
AT THAT CRITICAL NUMBER.
WELL IF IT'S CONCAVE UP
WE'D HAVE A RELATIVE MINIMUM AT X = 1.
SO LET'S WRITE THAT AGAIN.
WE HAVE A RELATIVE MINIMUM OF 1 AT X = 1.
REMEMBER SOMETIMES YOU CAN EXPRESS RELATIVE EXTREMA
AS AN ORDERED PAIR.
SOMETIMES OTHER TEXTBOOKS WANT IT WRITTEN OUT LIKE THIS.
LET'S GO AHEAD AND VERIFY WITH A GRAPH.
HERE'S THE POINT 1/1
AND WE CAN SEE THAT IT IS A RELATIVE MINIMUM
OR A LOW POINT ON THIS GRAPH.
LET'S GO AHEAD AND TRY ANOTHER EXAMPLE.
WE'LL START BY FINDING THE CRITICAL NUMBERS
SO LETS FIND THE FIRST DERIVATIVE 2 COSINE X.
NOW HERE WE HAVE TO APPLY THE CHAIN RULE.
THAT WOULD BE -SINE 2X x 2.
THIS IS ALWAYS DEFINED ON THIS INTERVAL
SO LET'S REWRITE THIS AND SEE IF WE CAN SET IT EQUAL TO ZERO
AND SOLVE.
WE HAVE 2 COSINE X - 2 SINE 2X = 0.
NOW WE HAVE A LITTLE BIT OF A PROBLEM HERE
BECAUSE WE HAVE TWO DIFFERENT FUNCTIONS
AND THIS ONE'S A DOUBLE ANGLE.
BUT IF YOU REMEMBER SINE 2X IS EQUAL TO 2 SINE X COSINE X.
SO WE'LL USE THIS IDENTITY AND DO A SUBSTITUTION HERE
AND THAT WOULD GIVE US 2 COSINE X - 2 x 2 SINE X COSINE X
OR 4 SINE X COSINE X,
AND THE REASON THAT'S HELPFUL IS NOW WE CAN FACTOR OUT 2 COSINE X
AND WE'RE LEFT WITH 1 - 2 SINE X = 0.
SO WE WANT TO KNOW WHEN COSINE X = 0
AND ALSO WHEN 1 - 2 SINE X = 0.
NOW WE'RE ONLY CONSIDERING THIS INTERVAL.
COSINE X = 0 WHEN X = PI/2.
LET'S GO AHEAD AND SOLVE THIS DIVIDE BY -2
THAT WOULD GIVE US SINE X = 1/2.
WELL THAT WOULD BE A 30 DEGREE REFERENCE ANGLE
IN THE FIRST AND SECOND QUADRANT.
SO THAT WOULD GIVE US X = PI/6 AND X = 5 PI/6.
YOU MAY WANT TO VERIFY THESE
BUT THESE ARE OUR THREE CRITICAL NUMBERS IN THIS GIVEN INTERVAL.
SO NOW WE HAVE TO FIND THE CORRESPONDING Y COORDINATE
OF THESE CRITICAL NUMBERS
BY EVALUATING THE FUNCTION AT THESE X VALUES.
LET'S GO AHEAD AND GO TO THE NEXT SCREEN AND DO THAT.
I'LL GO AHEAD AND PUT THESE IN ORDER.
WE KNOW THAT THE SINE OF PI/6 IS EQUAL TO 1/2 SO 2 x 1/2
THAT WOULD EQUAL 1
+ THE COSINE OF 2 x PI/6 THAT WOULD BE PI/3
IS EQUAL TO 1/2.
SO WE'D HAVE 3/2 FOR THE CORRESPONDING Y VALUE.
2 x THE SINE OF PI THAT WOULD BE 2 x 1 THAT WOULD GIVE US 2
+ THE COSINE OF PI IS EQUAL TO -1.
SO THAT WOULD GIVE US A VALUE OF 1 WHEN X IS PI/2
AND FOR F OF 5 PI/6
WE'RE GOING TO HAVE 1 + 1/2 OR 3/2 AGAIN.
OKAY. NOW WE HAVE TO DETERMINE
THE SINE OF THE SECOND DERIVATIVE
AT EACH OF THESE CRITICAL NUMBERS.
NOW I DIDN'T SHOW HOW TO FIND THIS SECOND DERIVATIVE.
LET'S GO AHEAD AND DO THAT NOW.
TO GO BACK TO THE PREVIOUS SCREEN
WE'RE GOING TO USE THIS FORM OF THE FIRST DERIVATIVE
TO DETERMINE THE SECOND DERIVATIVE.
SO WE HAVE F PRIME OF X = 2 COSINE X - 2 SINE 2X.
SO THE DERIVATIVE OF THIS OR THE SECOND DERIVATIVE
IS GOING TO EQUAL -2 SINE X - 2 x THE DERIVATIVE OF SINE 2X
WILL BE COSINE 2X x 2.
SO WE HAVE -2 SINE X - 4 COSINE 2X.
SO THAT'S WHERE THE SECOND DERIVATIVE CAME FROM.
NOW WE'LL ENTER THIS INTO THE CALCULATOR
AND THE DETERMINE THE SIGN OF THE SECOND DERIVATIVE
AT THESE CRITICAL NUMBERS.
MAKE SURE YOU'RE IN RADIAN MODE. I KNOW I ALREADY AM.
SO WE'LL GO AHEAD NOW AND ENTER IN OUR CRITICAL NUMBERS.
THE FIRST ONE IS PI/6.
SECOND DERIVATIVE IS NEGATIVE AND THEN WE'LL TYPE IN PI/2
THAT'S POSITIVE
AND THEN 5 PI/6
AND THAT'S NEGATIVE.
SO IT'S NEGATIVE, POSITIVE AND THEN NEGATIVE.
SO THAT MEANS THE FUNCTION'S CONCAVE DOWN
SO THAT WE'D HAVE A RELATIVE MAX.
HERE IT'S CONCAVE UP SO RELATIVE MIN,
CONCAVE DOWN ANOTHER RELATIVE MAX.
OKAY. SO WE'LL LEAVE IT AT THAT.
LET'S GO AHEAD AND VERIFY THIS WITH A GRAPH.
NOTICE WE HAVE 2 RELATIVE MAXIMUMS OF 3/2 AT PI/6
AND 5 PI/6.
WE HAVE A RELATIVE MINIMUM OF 1 AT PI/2.
THIS VERIFIES OUR WORK
AND THIS IS HOW YOU CAN USE THE SECOND DERIVATIVE
TO DETERMINE RELATIVE EXTREMA.
THANK YOU FOR WATCHING.