 – WE WANT TO DETERMINE
THE DOMAIN OF THE FUNCTION F(X)=THE SQUARE ROOT
OF THE QUANTITY X – 2 DIVIDED BY THE QUANTITY X – 5. FOR THIS FUNCTION,
BOTH THE NUMERATOR AND DENOMINATOR WILL GIVE
US RESTRICTIONS ON THE DOMAIN. LET’S START BY CONSIDERING
THE DENOMINATOR. WE KNOW DIVISION BY 0
IS UNDEFINED, SO ANY X VALUE THAT MAKES THE DENOMINATOR
=TO 0 MUST BE EXCLUDED. SO TO SHOW THIS,
LET’S SET X – 5=TO 0. WE SOLVE THIS, WE CAN SEE X=5
WOULD MAKE THE DENOMINATOR=TO 0 AND THEREFORE
WE MUST EXCLUDE THIS VALUE BECAUSE THAT’S WHEN
WE’D HAVE DIVISION BY 0. SO X CANNOT=5
IS ONE RESTRICTION ON THE DOMAIN. NOW LET’S CONSIDER
THE NUMERATOR. IN ORDER FOR THE NUMERATOR
TO BE A REAL NUMBER, THE NUMBER UNDERNEATH
THE SQUARE ROOT CANNOT BE NEGATIVE,
MEANING IT MUST BE>THAN OR=TO 0. SO THE NEXT RESTRICTION IS,
X – 2 MUST BE>THAN OR=TO 0, AGAIN, IN ORDER
FOR THE NUMERATOR TO BE A REAL NUMBER. SO IF WE SOLVE THIS FOR X,
WE HAVE X IS>THAN OR=TO +2. THIS IS THE SECOND RESTRICTION
ON OUR DOMAIN. SO THERE ARE SEVERAL WAYS
TO EXPRESS THE DOMAIN. ONE WAY WOULD BE TO SAY
THE DOMAIN IS WHEN X IS>THAN OR=TO +2
AND X DOES NOT=5. BUT LET’S ALSO GRAPH
THIS DOMAIN. SO WE’LL START BY LABELING
TO EXCLUDE 5 FROM THE DOMAIN, WE’LL MAKE AN OPEN POINT
ON THE NUMBER LINE AT +5. AND THEN WE’LL GRAPH
THE INEQUALITY X>THAN OR=TO +2,
KEEPING IN MIND X CAN’T=5. SO WE’D HAVE A CLOSED POINT
ON +2 AND THEN AN ARROW TO THE RIGHT. BUT BECAUSE WE HAVE
TO EXCLUDE 5, WE WOULD DRAW AN ARROW FROM HERE
TO THE OPEN CIRCLE ON 5 AND THEN CONTINUE
IT TO THE RIGHT OF +5. THIS WOULD BE THE GRAPH
OF THE DOMAIN OF THE FUNCTION. LET’S ALSO LIST
THE DOMAIN OF THE FUNCTION USING INTERVAL NOTATION. TO HELP US DO THIS, WE KNOW
AS WE CONTINUE RIGHT, WE’RE GOING TO APPROACH
POSITIVE INFINITY. SO WE HAVE TWO INTERVALS HERE. WE HAVE INTERVAL
FROM 2 TO 5 WHERE THE INTERVAL
IS CLOSED ON 2 AND OPEN ON 5. SO WE’D HAVE
A SQUARE BRACKET ON 2 AND A PARENTHESIS ON +5 UNION. AND THEN WE HAVE THE INTERVAL
FROM 5 TO INFINITY OPEN ON 5. SO PARENTHESIS 5, INFINITY. SO WE’VE GIVEN THE DOMAIN
THREE DIFFERENT WAYS. WE’VE EXPRESSED
IT USING INEQUALITIES, WE’VE EXPRESSED IT USING
INTERVAL NOTATION AND WE’VE EXPRESSED
IT GRAPHICALLY. AS A LAST CHECK FOR THE DOMAIN,
IT’S OFTEN HELPFUL TO GRAPH THE FUNCTION AND PROJECT
THE FUNCTION ONTO THE X-AXIS. SO HERE’S THE GRAPH
OF OUR FUNCTION. IF WE WERE TO PROJECT
THIS DOWN ONTO THE X-AXIS, OR YOU COULD THINK OF IT AS
MAKING A SHADOW ON THE X-AXIS, NOTICE HOW THE GRAPH
STARTS AT +2. THERE’S A BREAK AT +5,
BUT IT IS CONTINUOUS BETWEEN 2 AND 5
AND THEN TO THE RIGHT, A +5 IT’S CONTINUOUS. VERIFYING OUR DOMAIN IS CORRECT. I HOPE YOU FOUND THIS HELPFUL.

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