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- GIVEN F OF X = THE QUANTITY X + 2 SQUARED,
WE WANT TO DETERMINE THE DOMAIN SO F OF X IS INCREASING
AND 1 TO 1.
WE ALSO WANT TO GIVE THE RANGE AND USE INTERVAL NOTATION.
SO HERE'S A GRAPH OF OUR FUNCTION F OF X.
NOTICE IF WE DON'T RESTRICT THE DOMAIN,
THIS FUNCTION IS NOT ONE TO ONE
BECAUSE HORIZONTAL LINES WOULD INTERSECT THIS GRAPH
IN MORE THAN ONE POINT.
NOTICE THE FUNCTION IS ALSO DECREASING ON THE LEFT
AND INCREASING ON THE RIGHT.
SO NOTICE IF WE CONSIDER THIS FUNCTION
ONLY FROM THE VERTEX TO THE RIGHT,
THE FUNCTION IS INCREASING
AND IT'S ALSO 1 TO 1 BECAUSE HORIZONTAL LINES
WOULD ONLY INTERSECT THIS HALF OF THE GRAPH AT ONE POINT.
SO NOW WE'LL DETERMINE THE DOMAIN AND RANGE
IF WE ONLY WANT THIS HALF OF THE GRAPH.
WELL, THE DOMAIN IS A SET OF ALL POSSIBLE X VALUES,
SO IF WE PROJECT THIS GRAPH UNDER THE X AXIS,
NOTICE HOW THE DOMAIN WOULD BE FROM -2 TO THE RIGHT
OR FROM -2 TO INFINITY.
AND IT WOULD INCLUDE THE VERTEX,
SO WE'LL INCLUDE -2 IN THE DOMAIN.
SO THE DOMAIN, USING INTERVAL NOTATION,
WOULD BE FROM -2 TO INFINITY.
AND IT'S CLOSED ON -2 MEANING IT INCLUDES -2.
WE COULD ALSO EXPRESS THIS USING INEQUALITIES
AS X IS GREATER THAN OR = TO -2.
NOW LET'S CONSIDER THE RANGE.
THE RANGE IS A SET OF ALL POSSIBLE Y VALUES
OR OUTPUTS OF THIS FUNCTION ON THE RESTRICTED DOMAIN.
WELL, IF WE PROJECT THIS GRAPH ON TO THE Y AXIS,
NOTICE HOW THE SMALLEST Y VALUE WOULD BE 0,
AND FROM THERE IT INCREASES UPWARD TOWARD POSITIVE INFINITY.
NOW THE RANGE WOULD BE THE INTERVAL FROM 0 TO INFINITY,
CLOSED ON 0 MEANING IT INCLUDES 0,
OR WE COULD SAY Y IS GREATER THAN OR = TO 0.
WITH THIS RESTRICTION, THE FUNCTION F IS NOW ONE TO ONE,
SO WE CAN FIND F INVERSE OF X.
TO DO THIS, LET'S FIRST WRITE THE ORIGINAL FUNCTION
REPLACING F OF X WITH Y,
SO WE'D HAVE Y = THE QUANTITY X + 2 SQUARED.
AND THEN TO FIND THE INVERSE,
WE INTERCHANGE THE X AND Y VARIABLES,
AND THEN SOLVE FOR Y.
SO WE HAVE X = QUANTITY Y + 2 SQUARED.
AND NOW WE'LL SOLVE THIS FOR Y.
THE FIRST STEP WE'LL TAKE THE SQUARE ROOT
OF BOTH SIDES OF THIS EQUATION.
SO WE'D HAVE THE SQUARE ROOT OF X EQUALS
THE SQUARE ROOT OF THE QUANTITY Y + 2 SQUARED.
SO WE HAVE THE SQUARE ROOT OF X EQUALS--
NORMALLY THIS WOULD BE THE OPPOSITE VALUE OF Y + 2,
BUT BECAUSE OF THE RESTRICTIONS HERE
WE DON'T HAVE TO WORRY ABOUT THAT.
THIS WOULD JUST BE Y + 2.
LAST STEP WE'LL SUBTRACT 2 ON BOTH SIDES,
WE HAVE THE SQUARE ROOT OF X - 2 = Y.
THIS IS OUR INVERSE FUNCTION SOLVE FOR Y.
SO WE'LL GO AHEAD AND REPLACE Y WITH F INVERSE OF X.
F INVERSE OF X IS EQUAL TO THE SQUARE ROOT OF X - 2.
WE'RE ALSO ASKED TO GIVE THE DOMAIN AND RANGE.
LET'S GO AHEAD AND DO THAT.
BECAUSE THIS IS THE INVERSE OF FUNCTION F,
THE DOMAIN OF F IS GOING TO BE THE RANGE OF F INVERSE,
AND THE RANGE OF F WILL BE THE DOMAIN OF F INVERSE.
SO THE DOMAIN WILL BE FROM 0 TO INFINITY, CLOSED ON 0,
AND THE RANGE WILL BE THE INTERVAL FROM -2 TO INFINITY.
LET'S GO AHEAD AND FINISH BY VERIFYING THIS GRAPHICALLY.
WE KNOW THAT IF WE GRAPH FUNCTION F
ON THE RESTRICTED DOMAIN
AND WE GRAPH THE INVERSE FUNCTION ON ITS DOMAIN.
THE TWO FUNCTIONS SHOULD BE SYMMETRICAL
ACROSS THE LINE Y = X.
AND HERE'S THE GRAPH OF THE ORIGINAL FUNCTION
ON THE RESTRICTED DOMAIN,
AND HERE'S THE GRAPH OF THE INVERSE FUNCTION
GRAPHED OVER ITS DOMAIN.
NOW WE CAN SEE THAT THESE TWO GRAPHS ARE SYMMETRICAL
ACROSS THE LINE Y = X.
OKAY, I HOPE YOU FOUND THIS HELPFUL.