 – GIVEN F OF X, Y=THE SQUARE
ROOT OF THE QUANTITY Y – X, WHICH OF THE FOLLOWING REGIONS
ARE IN THE DOMAIN OF F OF X, Y? THE DOMAIN OF A FUNCTION
OF TWO VARIABLES IS A SET OF ALL POSSIBLE INPUTS OR SET OF ALL POSSIBLE ORDERED
PAIRS, IN THIS CASE,
ALL POSSIBLE VALUES OF X AND Y. SO BECAUSE WE HAVE
THE SQUARE ROOT HERE AND F OF X, Y
MUST BE A REAL NUMBER, THE RADICAND OR THIS EXPRESSION
UNDERNEATH THE SQUARE ROOT CAN’T BE NEGATIVE, WHICH MEANS
IN ORDER TO FIND THE DOMAIN, Y – X MUST BE GREATER THAN
OR EQUAL TO 0. IT’D BE OKAY TO EQUAL 0, BECAUSE
THE SQUARE ROOT OF 0 IS 0, THIS JUST CAN’T BE NEGATIVE. SO TO SOLVE THIS FOR A Y,
WE WOULD ADD X TO BOTH SIDES, GIVING US Y IS GREATER THAN
OR EQUAL TO X. SO IF WE GRAPH THIS LINEAR
INEQUALITY ON THE 2-DIMENSIONAL
COORDINATE PLANE, WE’LL KNOW THE REGION THAT FORMS
A DOMAIN OF F OF X, Y. SO FIRST WE’LL GRAPH THE LINE
Y=X AND THEN WE’LL DETERMINE
WHERE TO SHADE. SO IF WE’RE GOING
TO GRAPH Y=X, THIS IS IN SLOPED INTERCEPT
FORM WITH A FORM Y=MX + B WHERE THE SLOPE WOULD BE
THE COEFFICIENT OF X OR THE SLOPE IS 1 OR 1/1
AND THE Y INTERCEPT, BECAUSE THERE’S NO CONSTANT
WOULD BE 0. SO NOTICE HOW THE LINE WITH A Y
INTERCEPT OF 0 AND A SLOPE OF 1 WOULD BE THIS LINE HERE. AND BECAUSE Y MUST BE GREATER
THAN OR EQUAL TO X, THAT MEANS WE’RE GOING TO SHADE
ABOVE THE LINE, SO WE WOULD SHADE
THESE TWO REGIONS HERE. SO THIS REGION HERE INCLUDING
THE LINE MAKES UP THE DOMAIN
OF F OF X, Y, AND THEREFORE OUR ANSWER IS “A”
AND D. LET’S ALSO GRAPH
THIS 3-DIMENSIONAL SURFACE AND LOOK DOWN ON THE X, Y PLANE
TO VERIFY THIS DOMAIN. SO HERE’S THE GRAPH
OF OUR SURFACE. THIS WOULD BE THE +X-AXIS HERE,
AND THIS WOULD BE THE +Y-AXIS. SO LET’S ROTATE THIS
SO THAT WE HAVE THE +X-AXIS MOVING TOWARD THE RIGHT AND THE +Y-AXIS MOVING UPWARD. SO NOW IF WE LOOK DOWN
JUST THE X, Y PLANE– LIKE THIS, WE CAN SEE THAT THIS DOES VERIFY
THAT OUR DOMAIN IS CORRECT. THE DOMAIN CONSISTS OF
THIS REGION HERE WHICH WOULD BE WHEN Y IS
GREATER THAN OR EQUAL TO X. I HOPE YOU FOUND THIS HELPFUL.