- WELCOME TO A VIDEO THAT WILL INTRODUCE FUNCTIONS
OF TWO VARIABLES.
THE GOALS OF THE VIDEO ARE TO DETERMINE FUNCTION VALUES
OF FUNCTIONS OF TWO VARIABLES
AND ALSO TO DETERMINE THE DOMAIN OF A FUNCTION OF TWO VARIABLES.
THE FUNCTION Z = F OF XY IS A FUNCTION OF 2 VARIABLES.
SO Z IS DEFINED IN TERMS OF BOTH X AND Y
THEREFORE THE DOMAIN IS THE SET OF ALL POSSIBLE VALUES
OF X AND Y.
AND AGAIN X AND Y WILL BE THE INPUTS INTO THE FUNCTION
AND Z WILL BE THE OUTPUT.
SO THE RANGE IS A SET OF ALL POSSIBLE VALUES OF Z
FROM THE DOMAIN.
AND SINCE THE FUNCTION CONTAINS THREE VARIABLES X,Y,Z
THE GRAPH IS THE SET OF ALL POSSIBLE POINTS
CALLED ORDER TRIPLES X,Y,Z THAT ARE PRODUCED BY THE FUNCTION.
AND GEOMETRICALLY THE GRAPH IS A SURFACE IN SPACE.
SO LET'S FIRST START BY TAKING A LOOK AT SOME COOL GRAPHS
OF FUNCTIONS IN TERMS OF TWO VARIABLES.
HERE WE SEE Z = COSIGN X TIMES SIGN Y.
NOTICE IT PRODUCES A SURFACE WITH HILLS AND VALLEYS.
Z = THE NATURAL LOG OF 1 - X - Y, PRODUCES THIS SURFACE.
HERE WE HAVE Z = THE SQUARE ROOT OF 4 - X SQUARED - Y SQUARED.
AND THEN LASTLY WE HAVE Z = E THE POWER OF 1 - X SQUARED
MINUS Y SQUARED
WHICH LOOKS LIKE A PEAK.
LET'S TAKE A LOOK AT THESE IN ANOTHER WAY.
AGAIN, HERE'S THE GRAPH OF Z = COSIGN X TIMES SIGN Y.
HERE'S THE GRAPH OF Z = NATURAL LOG OF 1 - X - Y.
NOW THESE GRAPHS ARE DONE ON MAPLE
AND NOTICE AS THIS GRAPH MOVES DOWNWARD
MAPLE HAS A DIFFICULT TIME PLOTTING THE SURFACE
AND THAT CAN SOMETIMES HAPPEN ON SOME GRAPHING SOFTWARE.
HERE'S Z = THE SQUARE ROOT OF 4 - X SQUARED - Y SQUARE,
AGAIN HALF OF A SPHERE
AND THEN LASTLY WE HAVE Z =
EITHER THE POWER OF 1 - X SQUARED - Y SQUARED.
LET'S GO AND DETERMINE SOME FUNCTION VALUES.
NOTICE IN ORDER TO EVALUATE THIS FUNCTION
WE NEED BOTH AN X AND A Y VALUE WHICH ARE BOTH GIVEN.
SO WE'RE GONNA REPLACE X WITH NEGATIVE 3 AND Y WITH 3PI/2.
AND NOW WE'LL JUST EVALUATE THIS.
WE'RE GONNA HAVE NEGATIVE THREE TIMES THE COSIGN OF 3PI/2
IS = TO 0, SO THIS FUNCTION VALUE IS 0.
WHEN X IS -3 AND Y IS 3PI/2 THE Z COORDINATE WOULD BE 0.
FOR PART B WE HAVE X = 2 AND Y = PI/3.
SO WE'LL HAVE 2 X COSIGN PI/3.
COSIGN PI/3 IS THE SAME AS COSIGN 60 DEGREES,
SO THAT'S GONNA GIVE US 1/2,
WELL 2 TIMES 1/2 WOULD BE = TO 1.
SO WHEN X IS 2 AND WHY IS PI/3 THE Z COORDINATE WOULD BE 1.
IF WE WANTED TO WE COULD WRITE THIS AS AN ORDERED TRIPLE
THAT WOULD LOOK LIKE THIS, 2 PI/3, 1
AND FOR 2 IT'S A SIMILAR PROCESS.
WE'LL BASICALLY PERFORM SUBSTITUTION WHEN X IS 5
AND Y IS 3.
SO WE HAVE 5 SQUARED + 3 DIVIDED BY 3.
SO WE HAVE WHAT, 25 + 28 DIVIDED BY 3.
THAT DOESN'T SIMPLIFY
AND AGAIN IF WE WANTED TO THE ORDER TRIPLE WOULD BE 5 3 28/3.
AND LET'S GO AHEAD AND TAKE A LOOK AT 1 MORE, F OF -44,
SO WE GOTTA BE CAREFUL WITH OUR SIGNS HERE.
WE'RE GONNA SQUARE -4 AND THEN ADD 4 DIVIDED BY +4.
SO WE'RE GONNA HAVE POSITIVE 16 + 4 = 20 DIVIDED BY 4
IT'S GONNA GIVE US 5.
SO THE ORDER TRIPLE WOULD BE -4
4 5 WHICH REPRESENTS 1 POINT ON THIS SURFACE.
NOW, LET'S TALK ABOUT DETERMINING A DOMAIN
OF A FUNCTION IN TERMS OF 2 VARIABLES.
WELL, IF THIS FUNCTION HAS 2 INPUTS, BOTH X AND Y,
ALL THE POSSIBLE VALUES OF X AND Y WILL MAKE UP THE DOMAIN.
AND THE MORE THAT WE CAN REMEMBER
ABOUT OUR BASIC FUNCTIONS LIKE Y = SIGN X AS WE SEE HERE
AND Y = COSIGN X AS WE SEE HERE
WILL HELP US DETERMINE THE DOMAIN OF THIS FUNCTION
AND 2 VARIABLES.
WHAT I MEAN BY THAT IS FOR BOTH OF THESE FUNCTIONS DOWN BELOW
X CAN BE ANY REAL NUMBER.
SO LOOKING AT THIS FUNCTION HERE, X CAN BE ANY REAL NUMBER
AND SO CAN Y.
SO THAT TELLS US THE DOMAIN IS GOING TO BE ALL REAL NUMBERS
FOR BOTH X AND Y.
LET'S GO AND TAKE A LOOK AT ANOTHER ONE.
HERE WE HAVE Z = THE NATURAL LOG OF 1 - X - Y.
NOW, LOOKING AT THIS FUNCTION HERE AND TWO VARIABLES
SHOULD REMIND US OF THE FUNCTION AND 1 VARIABLE
OF Y = NATURAL LOG X.
IF YOU REMEMBER HERE'S THE GRAPH.
THE DOMAIN OF THIS FUNCTION WOULD BE X GREATER THAN 0
AND THE REASON THAT'S HELPFUL
IS IF YOU TAKE A LOOK AT THE GIVEN FUNCTION
AND TWO VARIABLES,
THE ARGUMENT 1 - X - Y MUST BE GREATER THAN 0
FOR THE SAME REASON WE HAVE A DOMAIN OF X IS GREATER THAN 0
HERE.
SO AGAIN THE DOMAIN WOULD BE 1 1 - X - Y IS GREATER THAN 0.
NOW, THIS IS THE DOMAIN BUT IT'S OFTEN BETTER TO MANIPULATE THIS
TO MAKE IT LOOK LIKE A MORE FAMILIAR FORM.
WHAT I MEAN BY THAT IS WE CAN MOVE THE X AND THE Y TERM
TO THE RIGHT SIDE.
SO WE WOULD HAVE 1 GREATER THAN X + Y.
FROM HERE'S IT'S PROBABLY BETTER TO GO AHEAD
AND SWITCH THIS AROUND
TO STATE THAT X + Y WOULD BE LESS THAN ONE.
AND OFTEN WHEN THE DOMAIN IS EXPRESSED
IN TERMS OF TWO VARIABLES
IT'S HELPFUL IF WE GRAPH THIS ON THE X,Y PLANE
TO SHOW THE REGION OF THE DOMAIN.
LET'S GO AHEAD AND DO THAT.
SO X + Y LESS THAN 1 IS GONNA BE A DASHED LINE
WITH AN X AND Y INTERCEPT OF +1.
THEN IF WE TEST THE ORIGIN 0,0 IN THIS INEQUALITY
THE 0 LESS THAN 1 IS TRUE
THEREFORE WE SHADE THIS HALF OF THE PLANE.
SO ANY POINT IN THIS REGION
WOULD BE PART OF THE DOMAIN OF THE GIVEN FUNCTION
IN TERMS OF 2 VARIABLES.
LET'S GO AND TAKE A LOOK AT ONE MORE.
HERE WE HAVE Z = THE SQUARE ROOT OF 4 - X SQUARED - Y SQUARED.
LOOKING AT THIS WE SHOULD BE REMINDED OF THE FUNCTION Y
= THE SQUARE ROOT OF X
WHEN WE HAVE A FUNCTION IN TERMS OF 1 VARIABLE.
AND AGAIN WE CAN SEE FROM THIS GRAPH
THE DOMAIN OF THIS FUNCTION WOULD BE
WHEN X IS GREATER THAN OR = TO 0.
REMEMBER, THE SQUARE ROOT OF 0 IS DEFINED.
SO WE'RE RELATED THIS DOMAIN TO THE GIVEN FUNCTION.
THIS SHOULD HELP US DETERMINE THAT 4 - X SQUARED - Y SQUARED
MUST BE GREATER THAN OR EQUAL TO ZERO
AND THAT WOULD BE OUR DOMAIN.
BUT NORMALLY IT'S NOT LEFT IN THIS FORM.
LET'S GO AHEAD AND MOVE THE X SQUARED AND Y SQUARED TERMS
TO THE OTHER SIDE
AND THAT WOULD GIVE US 4 GREATER THAN OR = TO
X SQUARED + Y SQUARED.
AND NOW WE'LL SWITCH THIS AROUND
MEANING WE HAVE X SQUARED + Y SQUARED
MUST BE LESS THAN OR = TO 4.
AND AGAIN LET'S GO AHEAD AND GRAPH THIS DOMAIN
ON THE X,Y PLANE.
NOTICE IF WE GRABBED X SQUARE + Y SQUARED = 4
THAT WOULD BE A CIRCLE CENTERED AT THE ORIGIN
WITH A RADIUS OF TWO.
AND SINCE IT CAN BE LESS THAN OR = TO FOUR
WE'LL MAKE A SOLID CIRCLE.
AGAIN IF WE TEST THE POINT 0 0,
0 IS LESS THAN OR = TO 4 IS TRUE.
SO FOR THIS PROBLEM THE DOMAIN WOULD BE ALL THE VALUES
ON THE CIRCLE OR INSIDE THE CIRCLE.
THIS GRAPH REPRESENTS THE DOMAIN OF THE GIVEN FUNCTION
AND TWO VARIABLES.
OKAY. THAT'S GONNA DO IT FOR THIS INTRODUCTORY VIDEO.
THANK YOU FOR WATCHING.