 – GIVEN F OF X, Y=
THE SQUARE ROOT OF THE QUANTITY X SQUARED – Y SQUARED, WHICH OF THE FOLLOWING REGIONS
ARE IN THE DOMAIN OF F OF X, Y WHERE THE DOMAIN OF A FUNCTION
OF TWO VARIABLES IS THE SET OF ALL POSSIBLE
INPUTS OR SET OF ALL POSSIBLE
ORDERED PAIRS? SO IN THIS CASE, IT’D BE ALL
POSSIBLE VALUES OF X AND Y. BUT BECAUSE WE HAVE THIS
SQUARE ROOT HERE AND F OF X, Y
MUST BE A REAL NUMBER, WE KNOW THE RADICAND
OR X SQUARED – Y SQUARED CAN’T BE NEGATIVE, AND THEREFORE TO HELP US
FIND THE DOMAIN, WE’LL SOLVE INEQUALITY
X SQUARED – Y SQUARED MUST BE GREATER THAN
OR EQUAL TO 0. SO LET’S WORK ON SOLVING THIS
FOR Y. LET’S FIRST ADD Y SQUARED
TO BOTH SIDES WHICH WOULD GIVE US X SQUARED IS GREATER THAN OR EQUAL
TO Y SQUARED. IF WE PREFER TO HAVE
THE Y SQUARED ON THE LEFT, WE CAN SWITCH THIS AROUND. THIS IS EQUIVALENT TO Y SQUARED IS LESS THAN OR EQUAL TO
X SQUARED. NOTICE IN BOTH CASES, THE INEQUALITY SYMBOL IS
POINTING TOWARD THE Y SQUARED. AND NOW HERE’S WHERE
IT GET’S A LITTLE BIT TRICKY. WE WANT Y HERE, NOT Y SQUARED. BUT NOTICE IN BOTH CASES IF
WE’RE SQUARING X AND SQUARING Y, THE RESULT WOULD BE POSITIVE. SO WHEN WE TAKE THE SQUARE ROOT
OF BOTH SIDES HERE, THE SQUARE ROOT OF Y SQUARED
WOULD BE THE ABSOLUTE VALUE OF Y TO ASSURE THAT THIS RESULT
IS POSITIVE. AND THE SAME THING ON THE RIGHT, THE SQUARE ROOT OF X SQUARED IS EQUAL TO THE ABSOLUTE VALUE
OF X. SO TO DETERMINE THE DOMAIN
OF F OF X, Y, WE WANT TO GRAPH THIS INEQUALITY
ON THE COORDINATE PLANE. TO DO THIS THOUGH, WE’LL BREAK THIS UP INTO
TO SEPARATE INEQUALITIES BASED UPON THE SIGN OF Y. SO FOR EXAMPLE, IF Y IS GREATER
THAN OR EQUAL TO 0, WE DON’T NEED THE ABSOLUTE VALUE
AROUND Y, AND THEREFORE WE CAN SAY THAT WHEN Y IS GREATER THAN OR EQUAL
TO 0, Y WILL ALWAYS BE
LESS THAN OR EQUAL TO THE ABSOLUTE VALUE OF X. BUT IF Y IS NEGATIVE
OR LESS THAN 0, THIS RESULT STILL HAS TO BE
POSITIVE. SO IF Y IS NEGATIVE, WE WANT TO
DROP THE ABSOLUTE VALUE. WE’D HAVE TO TAKE THE OPPOSITE
OF Y, SO WE COULD SAY, -Y IS LESS THAN OR EQUAL TO
THE ABSOLUTE VALUE OF X. SO IF WE SOLVE THIS FOR Y,
WE’D MULTIPLY BY -1. WHEN WE DO THIS THOUGH, WE HAVE TO REVERSE
THE INEQUALITY SYMBOL. THIS WOULD BE EQUIVALENT TO Y IS GREATER THAN OR EQUAL TO –
ABSOLUTE VALUE OF X. AND THEREFORE TO FIND THE DOMAIN
OF F OF X, Y, WE WANT TO GRAPH THIS INEQUALITY
AND THIS INEQUALITY AND DETERMINE
WHEN THEY’RE BOTH TRUE. SO TO GRAPH Y IS LESS THAN
OR EQUAL TO THE ABSOLUTE VALUE OF X, WE SHOULD RECOGNIZE
THAT THIS V SHAPE HERE WOULD BE THE ABSOLUTE VALUE
OF X. SO IF WE WANT TO GRAPH Y IS LESS THAN OR EQUAL TO
THE ABSOLUTE VALUE OF X, WE WOULD ACTUALLY SHADE DOWN. BUT WE ALSO WANT TO KNOW WHEN Y IS GREATER THAN OR EQUAL
TO -ABSOLUTE VALUE OF X. WELL THIS V SHAPE HERE
THAT OPENS DOWN WOULD BE THE GRAPH OF Y=
– ABSOLUTE VALUE OF X. SO TO SHADE WHEN Y
IS GREATER THAN OR EQUAL TO -ABSOLUTE VALUE OF X, WE’D ACTUALLY SHADE UP. SO WE CAN SEE THESE INEQUALITIES
ARE BOTH TRUE IN THIS REGION AND IN THIS REGION, AND THEREFORE THE DOMAIN
OF F OF X, Y WOULD BE IN THE REGIONS B AND D. NOW, LET’S GRAPH THE SURFACE
IN 3D AND LOOK DOWN ON THE X, Y PLANE
TO VERIFY THIS DOMAIN. HERE’S THE GRAPH OF OUR SURFACE. THIS–HAVING A LITTLE DIFFICULT
TIME GRAPHING THE SURFACE HERE NEAR THE X, Y PLANE, BUT WE SHOULD BE ABLE TO TELL
IT’S HALF A CONE WHERE THIS AXIS HERE
IS THE +X- AXIS AND THIS AXIS HERE
WOULD BE THE +Y-AXIS. SO LET’S ROTATE THIS SO THAT THE
+X-AXIS POINTS TOWARD THE RIGHT AND THE +Y-AXIS POINTS UPWARD. NOW THAT WE HAVE THIS ROTATED, LET’S LOOK DOWN ON THE X, Y
PLANE TO VERIFY OUR DOMAIN. SO IF WE PAUSE HERE, THIS DOES VERIFY OUR DOMAIN IS WHEN Y IS LESS THAN OR EQUAL
TO THE ABSOLUTE VALUE OF X AND WHEN Y IS GREATER THAN
OR EQUAL TO THE -ABSOLUTE VALUE OF X OR THIS REGION AND THIS REGION
IN THE X, Y PLANE. I HOPE YOU FOUND THIS HELPFUL.