Ex:  Domain of a Quotient and Composite Functions

Ex: Domain of a Quotient and Composite Functions


– WE’RE ASKED TO FIND
THE DOMAIN OF F OF X DIVIDED BY G OF X,
F OF G OF X, AND G OF F OF X. WE ALSO WANT TO EXPRESS
THE DOMAIN USING INTERVAL NOTATION. SO WE’LL DO THESE ONE
AT A TIME. LET’S FIRST START
WITH THE QUOTIENT OF F AND G. SO F OF X DIVIDED BY G OF X, WELL F OF X IS 2 DIVIDED
BY THE SQUARED OF X AND G OF X
IS (X SQUARED – X) – 6. WE’RE GOING TO HAVE SEVERAL
RESTRICTIONS ON THE DOMAIN SO LET’S KEEP TRACK OF THOSE
OVER HERE ON THE SIDE. LET’S FIRST TALK
ABOUT THE RESTRICTIONS BASED UPON THIS SQUARED OF X
HERE. WE KNOW X CAN’T BE NEGATIVE BECAUSE THE SQUARED
OF NEGATIVE IS NOT REAL BUT NOTICE
IF X IS EQUAL TO ZERO WE’D ALSO HAVE DIVISION
BY ZERO. SO RIGHT AWAY WE KNOW THAT
X HAS TO BE GREATER THAN ZERO AND NOW LET’S LOOK
AT THE DENOMINATOR, WE KNOW (X SQUARED – X) – 6
CAN’T EQUAL ZERO BECAUSE IF IT DID
WE’D HAVE DIVISION BY ZERO. SO LET’S FIND THE VALUES OF X THAT WOULD MAKE
THE DENOMINATOR EQUAL TO ZERO AND THEN WE’LL EXCLUDE
THOSE VALUES AS WELL. WELL THIS FACTORS NICELY. THE FACTORS OF -6
THAT ADD TO -1 ARE -3 AND +2. SO THE VALUES OF X THAT MAKE
THIS PRODUCT EQUAL TO ZERO MUST BE EXCLUDED. WELL 3 WOULD MAKE X – 3=0
SO X CAN’T=3 AND -2 WOULD MAKE X + 2=0
SO X CAN’T ALSO EQUAL -2. SO LET’S SEE
IF WE CAN GRAPH OUR DOMAIN AND THEN WE’LL WRITE
THE DOMAIN USING INTERVAL NOTATION. LOOKING AT OUR RESTRICTIONS, WE KNOW X HAS TO BE GREATER
THAN ZERO BUT IT ALSO CAN’T EQUAL 3. SO WE’LL HAVE TO HAVE
AN OPEN CIRCLE ON 3, OPEN CIRCLE ON ZERO, SO WE’LL HAVE THE INTERVAL
FROM ZERO TO 3 AND THEN FROM 3
TO POSITIVE INFINITY. NOTICE WE DON’T HAVE TO WORRY
ABOUT X NOT EQUALING -2 BECAUSE -2 IS NOT GREATER
THAN ZERO. SO THE DOMAIN
OF OUR QUOTIENT FUNCTION WILL BE THE OPEN INTERVAL
FROM ZERO TO 3 AND THEN THE OPEN INTERVAL
FROM 3 TO INFINITY.   NOW LET’S FIND THE DOMAIN OF F
OF G OF X ON THE NEXT SLIDE, BEFORE WE FIND
OUR COMPOSITE FUNCTION LET’S REVIEW HOW TO FIND
THE DOMAIN OF A COMPOSITE FUNCTION. THE DOMAIN OF F OF G OF X MUST CONTAIN THE RESTRICTIONS
OF G OF X, THE INNER FUNCTION AND THE RESTRICTIONS
OF THE COMPOSITE FUNCTION. SO WE’LL START BY DETERMINING
THE RESTRICTIONS FROM THE INNER FUNCTION,
FIND THE COMPOSITE FUNCTION, AND THEN FIND ANY OTHER
ADDITIONAL RESTRICTIONS. SO F OF G OF X CAN BE WRITTEN
USING THIS NOTATION HERE. SO G OF X
IS OUR INNER FUNCTION. LET’S KEEP TRACK
OF OUR RESTRICTIONS OVER HERE ON THE SIDE. WELL G OF X
IS A QUADRATIC FUNCTION AND THE DOMAIN OF A QUADRATIC
FUNCTION IS ALL REAL NUMBERS. SO THERE’S NO RESTRICTIONS
WE HAVE TO WORRY ABOUT FOR THE INNER FUNCTION G OF X. SO WE’LL GO AHEAD AND FIND
OUR COMPOSITE FUNCTION AND SEE IF THERE ARE
ANY OTHER RESTRICTIONS. SO TO FIND OUR COMPOSITE
FUNCTION F OF G OF X WE’RE GOING TO REPLACE G OF X
WITH (X SQUARED – X) – 6 WHICH WILL GIVE US
F OF X SQUARED – X – 6 AND THIS BECOMES THE INPUT
INTO FUNCTION F SO WE’LL SUBSTITUTE THIS
QUANTITY IN FOR THIS X HERE. SO WE’LL HAVE 2 DIVIDED BY THE SQUARE ROOT
OF X SQUARED – X – 6. SO THE DOMAIN
OF OUR COMPOSITE FUNCTION WILL BE THE DOMAIN
OF THIS FUNCTION HERE SINCE THERE WERE
NO RESTRICTIONS FOR THE DOMAIN OF G OF X. WELL WE KNOW WE CAN’T HAVE
DIVISION BY ZERO AND WE ALSO KNOW
THAT THE NUMBER UNDERNEATH THE SQUARE ROOT
CAN’T BE NEGATIVE. SO THE RESTRICTION
THAT WE HAVE IS THAT (X SQUARED – X) – 6
MUST BE GREATER THAN ZERO. SO THE DOMAIN
WILL BE THE SOLUTION TO THIS QUADRATIC INEQUALITY. SO TO SOLVE THIS WE’LL DETERMINE WHAT X VALUES
MAKE THIS EQUAL TO ZERO AND THEN WE’LL FORM INTERVALS TO SEE WHICH INTERVALS
SATISFY THIS INEQUALITY. SO WE’RE GOING TO SOLVE
(X SQUARED – X) – 6=0. WHICH WE ALREADY FACTORED
ON THE PREVIOUS SLIDE, IT’S (X – 3) x (X + 2). SO X=3 AND X=-2
WILL BE THE VALUES THAT MAKE THIS EQUAL TO ZERO. SO NOW WE’LL DRAW
A NUMBER LINE BECAUSE THIS CAN’T EQUAL ZERO WE’LL MAKE OPEN CIRCLES
ON 3 AND -2. SO NOW WE’LL PICK TEST VALUES
IN EACH INTERVAL TO SEE WHICH SATISFY
THE ORIGINAL INEQUALITY. SO FOR THIS INTERVAL HERE
LET’S GO AHEAD AND TEST ZERO. WELL IF X IS ZERO WE’D HAVE
(0 – 0) – 6, -6 IS NOT GREATER THAN ZERO, THEREFORE THIS INTERVAL
IS FALSE. LET’S GO AHEAD AND CHECK
LET’S SAY X=4. WE’D HAVE 16 – 4
THAT’S 12 – 6 THAT’S +6. +6 IS GREATER THAN ZERO SO
THIS ENTIRE INTERVAL IS TRUE MEANING IT WOULD BE
PART OF OUR DOMAIN AND LET’S ALSO GO AHEAD
AND TEST LET’S SAY -3. -3 SQUARED IS 9, – -3
WOULD BE + 3 OR 12. 12 – 6=6
WHICH IS GREATER THAN ZERO. THEREFORE THE INTERVAL
ON THE LEFT IS ALSO PART OF OUR DOMAIN. SO THIS IS APPROACHING
POSITIVE INFINITY, THIS IS APPROACHING
NEGATIVE INFINITY. SO THE DOMAIN
OF OUR COMPOSITE FUNCTION WOULD BE
FROM NEGATIVE INFINITY TO -2 NOT INCLUDING -2 AS WELL AS THE INTERVAL
FROM 3 TO INFINITY, AND THE INTERVAL IS OPEN ON 3. AND NOW FOR THE LAST QUESTION, WE WANT TO FIND THE DOMAIN
OF G OF F OF X. WE’LL GO AHEAD AND WRITE THIS
USING THIS NOTATION HERE, AND BECAUSE WE HAVE
A COMPOSITE FUNCTION WE FIRST NEED TO CONSIDER
THE RESTRICTIONS ON THE DOMAIN OF THE INNER FUNCTION OR IN THIS CASE F OF X. LET’S GO AHEAD AND KEEP TRACK
OF OUR RESTRICTIONS. LET’S GO AHEAD AND TAKE A LOOK
AT OUR INNER FUNCTION F OF X. WE HAVE 2
DIVIDED BY THE SQUARED OF X BUT WE KNOW X
CAN’T BE NEGATIVE BUT IT ALSO CAN’T BE ZERO BECAUSE WE HAVE DIVISION
BY ZERO. WHICH MEANS X MUST BE GREATER
THAN ZERO TO BEGIN WITH. NOW WE’LL FIND
OUR COMPOSITE FUNCTION AND SEE IF THERE ARE MORE
RESTRICTIONS ON THE DOMAIN. SO THIS IS EQUAL TO G OF F
OF X IS EQUAL TO 2 DIVIDED BY THE SQUARED OF X AND THIS BECOMES
THE INPUT INTO FUNCTION G. SO WE’LL SUBSTITUTE 2
DIVIDED BY THE SQUARED OF X HERE AND HERE. SO WE’LL HAVE 2 DIVIDED BY
THE SQUARED OF X SQUARED – 2 DIVIDED BY THE SQUARED OF X
AND – 6. LET’S GO AHEAD
AND SIMPLIFY THIS. WE’D HAVE 4 DIVIDED BY X – 2 DIVIDED BY THE SQUARED
OF X – 6. WE’VE ALREADY SAID THAT X
HAS TO BE GREATER THAN ZERO. NOTICE OUR COMPOSITE FUNCTION DOES NOT ADD ANY ADDITIONAL
RESTRICTIONS FOR THE DOMAIN, AND THEREFORE OUR DOMAIN IS GOING TO BE X
IS GREATER THAN ZERO. SO IF WE WERE TO GRAPH THAT
OF COURSE IT WOULD LOOK LIKE THIS APPROACHING
POSITIVE INFINITY SO THE DOMAIN OF G OF F OF X
IS FROM ZERO TO INFINITY NOT INCLUDING ZERO. I REALIZE THESE PROBLEMS
CAN BE FAIRLY INVOLVED, SO I HOPE THESE EXPLANATIONS
WERE HELPFUL.  

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