# Ex: Restrict the Domain to Make a Function 1 to 1, Then Find the Inverse

– 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