Ex - Determine increasing/decreasing/concavity intervals of a rational function

- HERE WE'RE GIVEN A RATIONAL FUNCTION. IN THE NUMBER ONE, ASKED TO FIND THE DOMAIN, NUMBER TWO, ASKED TO DETERMINE THE OPEN INTERVALS FOR WHICH THE FUNCTION IS INCREASING OR DECREASING, AND THEN STEP THREE, FIND THE OPEN INTERVALS WHERE THE FUNCTION IS CONCAVE UP OR CONCAVE DOWN. LET'S BEGIN BY DETERMINING THE DOMAIN OF THE GIVEN FUNCTION. WHEN WORKING WITH RATIONAL FUNCTIONS, WE MUST EXCLUDE VALUES OF X WHERE WE'D HAVE DIVISION BY ZERO OR WHERE THE DENOMINATOR WOULD BE EQUAL TO ZERO. SO FOR NUMBER ONE, DEFINE THE DOMAIN, WE'LL DETERMINE WHICH VALUES OF X WE MUST EXCLUDE FROM THE DOMAIN BY SETTING THE DENOMINATOR EQUAL TO ZERO. SO IF WE SET 3X + 1 = 0 AND SOLVE, WE'D SUBTRACT 1 ON BOTH SIDES. SO WE'D HAVE 3X = -1, DIVIDE BY +3. SO X = -1/3 WOULD MAKE THE DENOMINATOR EQUAL TO ZERO, MEANING WE'D HAVE DIVISION BY ZERO, AND THEREFORE THE FUNCTION WOULD BE UNDEFINED. SO THE DOMAIN WOULD BE ALL REAL NUMBERS EXCEPT -1/3. SO USING INTERVAL NOTATION WE CAN SAY THE OPEN INTERVAL FROM NEGATIVE INFINITY TO -1/3 UNION THE OPEN INTERVAL FROM -1/3 TO POSITIVE INFINITY. SOME TEXTBOOKS ALSO JUST SAY X CAN'T EQUAL -1/3. OR WE COULD SAY ALL REAL NUMBERS EXCEPT X = -1/3. SO THERE ARE SEVERAL WAYS TO EXPRESS THE DOMAIN. AND NOW LETS DETERMINE THE INTERVAL FOR WHICH THE FUNCTION IS INCREASING OR DECREASING. TO DO THIS, WE'LL FIRST FIND THE CRITICAL NUMBERS OF THE GIVEN FUNCTION, WHICH WOULD BE WHERE THE FIRST DERIVATIVE IS EQUAL TO ZERO OR UNDEFINED. ONCE WE FIND THE CRITICAL NUMBERS WE'LL DIVIDE THE DOMAIN OF THE FUNCTION INTO INTERVALS AND TEST THE SIGN OF THE FIRST DERIVATIVE. IF THE FIRST DERIVATIVE IS POSITIVE THE FUNCTION IS INCREASING, IF THE FIRST DERIVATIVE IS NEGATIVE THE FUNCTION IS DECREASING. SO LET'S GO AHEAD AND FIND THE FIRST DERIVATIVE ON THE NEXT SLIDE. NOTICE WE'LL HAVE TO APPLY THE QUOTIENT RULE IN ORDER TO FIND OUR DERIVATIVE HERE. SO F PRIME OF X WILL BE EQUAL TO-- BEGINNING WITH THE DENOMINATOR, WHICH WOULD BE THE DENOMINATOR SQUARED, OUR NUMERATOR IS GOING TO BE THE DENOMINATOR TIMES THE DERIVATIVE OF THE NUMERATOR MINUS THE NUMERATOR TIMES THE DERIVATIVE OF THE DENOMINATOR. SO NOW WE'LL FIND THE DERIVATIVE HERE AND HERE AND SIMPLIFY. OUR DENOMINATOR IS GOING TO STAY THE SAME. DERIVATIVE OF 2X + 5 WOULD JUST BE 2. SO WE'LL HAVE 2 x THE QUANTITY 3X + 1 - THE DERIVATIVE OF 3X + 1 IS 3 SO WE'D HAVE - 3 x THE QUANTITY 2X + 5. LET'S GO AND SIMPLIFY THE NUMERATOR. IF WE DISTRIBUTE WE'D HAVE 6X + 2. HERE WE'LL DISTRIBUTE -3, SO WE'D HAVE -6X - 15. WELL, 6X - 6X IS 0 AND WE HAVE 2 - 15 THAT'S -13 DIVIDED BY THE QUANTITY 3X + 1 TO THE 2ND. AND, AGAIN, THE CRITICAL NUMBERS OCCUR WHERE THIS FUNCTION IS EQUAL TO 0 OR UNDEFINED. WELL, NOTICE HOW THE NUMERATOR IS A CONSTANT, SO THIS WILL NEVER EQUAL ZERO. BUT NOTICE HOW IT IS UNDEFINED WHEN WE HAVE DIVISION BY ZERO OR WHEN THE QUANTITY 3X + 1 = 0. WE SOLVE THIS FOR X JUST LIKE BEFORE, WE'D HAVE X = -1/3. THIS IS ALSO THE SAME VALUE WHERE THE FUNCTION IS UNDEFINED. AND WE DO HAVE TO CONSIDER THIS VALUE WHEN DETERMINING THE SIGN OF THE FIRST DERIVATIVE. IT COULD CHANGE SIGNS TO THE LEFT AND RIGHT OF -1/3. SO NOW LET'S GO AHEAD AND SET UP THE INTERVALS TO TEST THE SIGN OF THE FIRST DERIVATIVE. BECAUSE OUR DOMAIN WAS ALL REAL NUMBERS EXCEPT X = -1/3-- BECAUSE THE DOMAIN OF THE FUNCTION WAS ALL REAL NUMBERS EXCEPT -1/3 AND THE FIRST DERIVATIVE IS UNDEFINED AT -1/3, OUR TWO INTERVALS TO TEST WILL BE FROM NEGATIVE INFINITY TO -1/3, AND THE OPEN INTERVAL FROM -1/3 TO INFINITY. SO NOW WE'LL TEST THIS ON THE FIRST DERIVATIVE AND EACH OF THESE INTERVALS TO DETERMINE IF THE FUNCTION IS INCREASING OR DECREASING ON THAT INTERVAL. SO FOR THIS FIRST INTERVAL LET'S GO AHEAD AND TEST X = -1. SO F PRIME OF -1 WOULD BE -13 DIVIDED BY 3 x -1 + 1. WE DON'T HAVE TO FIND THE EXACT VALUE. WE JUST HAVE TO DETERMINE IF THIS IS GOING TO BE POSITIVE OR NEGATIVE. AND NOTICE HOW WE'D HAVE A NEGATIVE DIVIDED BY A POSITIVE VALUE, AND THEREFORE THIS IS GOING TO BE NEGATIVE OR LESS THAN ZERO. SO THE FIRST DERIVATIVE IS NEGATIVE OVER THIS ENTIRE INTERVAL, AND THEREFORE THE FUNCTION IS DECREASING OVER THIS ENTIRE INTERVAL. AND NOW FOR THE SECOND INTERVAL LET'S TEST X = 1. SO F PRIME OF 1 WOULD BE EQUAL TO, AGAIN, -13 DIVIDED BY THE QUANTITY 3 x 1 + 1 SQUARED. AGAIN, WE'D HAVE A NEGATIVE DIVIDED BY A POSITIVE, WHICH IS NEGATIVE OR LESS THAN ZERO, AND THEREFORE THE FIRST DERIVATIVE IS NEGATIVE OVER THIS ENTIRE INTERVAL. SO, AGAIN, THE FUNCTION IS DECREASING OVER THIS ENTIRE INTERVAL. WHICH MEANS, THE FUNCTION IS DECREASING OVER ITS ENTIRE DOMAIN. LET'S GO AHEAD AND CHECK THIS ON THE GRAPH. NOTICE THAT X = -1/3, WE HAVE A VERTICAL ASYMPTOTE.   AND ON THE LEFT ON THE INTERVAL FROM NEGATIVE INFINITY TO -1/3, NOTICE HOW THE FUNCTION IS DECREASING. AS X INCREASES Y DECREASES OR AS THEY MOVE FROM LEFT TO RIGHT WE'RE GOING DOWN-HILL. AND ALSO ON THE RIGHT SIDE OR ON THE OPEN INTERVAL FROM -1/3 TO POSITIVE INFINITY, AGAIN, AS X INCREASES THE Y VALUES DECREASE, OR AS WE MOVE FROM LEFT TO RIGHT THE FUNCTION IS GOING DOWN-HILL. SO THIS VERIFIES THE FUNCTION IS DECREASING OVER ITS ENTIRE DOMAIN. AND NOW FOR THE THIRD PART, WE WANT TO DETERMINE THE INTERVALS FOR WHICH THE FUNCTION IS CONCAVE UP OR DOWN. A FUNCTION IS CONCAVE UP WHEN THE SECOND DERIVATIVE IS POSITIVE, AND A FUNCTION IS CONCAVE DOWN WHEN THE SECOND DERIVATIVE IS NEGATIVE. SO TO DETERMINE THE CONCAVITY, WE BEGIN BY DETERMINING WHERE THE SECOND DERIVATIVE IS UNDEFINED OR EQUAL TO ZERO. SO TO FIND THE SECOND DERIVATIVE WE'LL FIND THE DERIVATIVE OF THE FIRST DERIVATIVE. SO THE FIRST DERIVATIVE IS EQUAL TO -13 DIVIDED BY THE QUANTITY 3X + 1 TO THE 2ND. LET'S GO AHEAD AND WRITE THIS AS -13 x THE QUANTITY 3X + 1 RAISED TO THE POWER OF -2. THIS WAY WE DON'T HAVE TO APPLY THE QUOTIENT RULE AGAIN. WE CAN JUST APPLY THE EXTENDED POWER RULE. SO THE SECOND DERIVATIVE FUNCTION WOULD BE EQUAL TO-- WE DO HAVE TO APPLY THE CHAIN RULE HERE WHERE THE INNER FUNCTION 3X + 1 WE CAN LET BE EQUAL TO U. SO WE CAN THINK OF THIS AS -13U TO THE -2, WHICH MEANS TO FIND THE DERIVATIVE WE WOULD MULTIPLY BY -2 THAT WOULD BE 26. THEN WE'D HAVE U TO THE -3, WHICH IS REALLY THE QUANTITY 3X + 1 TO THE -3 x U PRIME, WHICH HERE THE DERIVATIVE OF 3X + 1 WOULD JUST BE 3. SO OUR SECOND DERIVATIVE IS GOING TO BE 26 x 3 THAT'S 78, DIVIDED BY THE QUANTITY 3X + 1 TO THE 3 POWER. IF WE MOVE THIS TO THE DENOMINATOR IT CHANGES THE SIGN OF THE EXPONENT. BUT ONCE AGAIN, NOTICE HOW THE SECOND DERIVATIVE IS NEVER GOING TO BE EQUAL TO ZERO BECAUSE WE HAVE A CONSTANT IN THE NUMERATOR. SO IT'S ONLY GOING TO BE UNDEFINED WHEN THE BASE OF 3X + 1 = 0. WELL, JUST AS BEFORE, 3X + 1 = 0 WHEN X = -1/3, WHICH IS NOT IN THE DOMAIN, BUT WE'LL STILL USE THIS VALUE TO SET UP THE INTERVALS TO TEST FOR CONCAVITY. THE FUNCTION MAY CHANGE CONCAVITY TO THE LEFT AND RIGHT OF -1/3. SO NOW USING THE SECOND DERIVATIVE, AND THE FACT THAT IT'S UNDEFINED AT X = -1/3, OUR TWO INTERVALS, ONCE AGAIN, WILL BE THE OPEN INTERVAL FROM NEGATIVE INFINITY TO -1/3 AND THE OPEN INTERVAL FROM -1/3 TO POSITIVE INFINITY. BUT NOW BECAUSE WE'RE TALKING ABOUT CONCAVITY, WE'LL BE TESTING THE SIGN OF THE SECOND DERIVATIVE NOT THE FIRST DERIVATIVE. SO FOR THIS FIRST INTERVAL, LET'S, AGAIN, TEST X = -1. SO F DOUBLE PRIME OF -1 WOULD BE 78 DIVIDED BY 3 x -1 + 1 TO THE 3RD. WELL, NOTICE HERE IN THE DENOMINATOR WE'D ACTUALLY HAVE -2 CUBED, SO WE'D HAVE 78 DIVIDED BY -8, WHICH IS OBVIOUSLY GOING TO BE NEGATIVE OR LESS THAN ZERO. AND THEREFORE, THE SECOND DERIVATIVE IS NEGATIVE ON THIS ENTIRE INTERVAL, WHICH MEANS THE FUNCTION WOULD HAVE TO BE CONCAVE DOWN OVER THIS ENTIRE INTERVAL. AND NOW FOR THE SECOND INTERVAL, LET'S TEST X = 1. SO F DOUBLE PRIME OF 1 WOULD BE 78 DIVIDED BY THE QUANTITY 3 x 1 + 1 CUBED. WELL, WE SHOULD BE ABLE TO RECOGNIZE THIS WILL BE A POSITIVE DIVIDED BY A POSITIVE, WHICH OF COURSE IS POSITIVE OR GREATER THAN ZERO. AND THEREFORE, THE SECOND DERIVATIVE IS POSITIVE OVER THIS INTERVAL, MEANING THE FUNCTION IS CONCAVE UP OVER THIS INTERVAL. LET'S GO AND CHECK THE CONCAVITY GRAPHICALLY AS WELL. REMEMBER FUNCTION IS CONCAVE UP. IT RESEMBLES THIS SHAPE. AND IF THE FUNCTION IS CONCAVE DOWN IT RESEMBLES THIS SHAPE. SO LOOKING AT OUR FUNCTION, NOTICE ON THE LEFT SIDE OR ON THE OPEN INTERVAL FROM NEGATIVE INFINITY TO -1/3 THIS PIECE HERE IS CONCAVE DOWN. IT RESEMBLES THIS PIECE HERE. AND THEN ON THE RIGHT, NOTICE HOW THIS FUNCTION IS CONCAVE UP, WHICH IS THE OPEN INTERVAL FROM -1/3 TO POSITIVE INFINITY. NOTICE HOW THIS PIECE RESEMBLES THIS PIECE HERE. I HOPE YOU FOUND THIS EXPLANATION HELPFUL.