 – WE’RE GIVEN F OF X EQUALS THE SQUARE ROOT OF THE QUANTITY
(2X – 1) – 3. WE WANT TO DETERMINE THE DOMAIN
AND RANGE OF THE GIVEN FUNCTION AND THEN
FIND THE INVERSE FUNCTION. BECAUSE OUR FUNCTION CONTAINS
A SQUARE ROOT IN ORDER FOR THE FUNCTION VALUE
TO BE REAL THE NUMBER UNDERNEATH
THE SQUARE ROOT OR THE RADICAND WHICH IN THIS CASE 2X – 1
CAN’T BE NEGATIVE WHICH MEANS 2X – 1 MUST BE
GREATER THAN OR EQUAL TO ZERO. SINCE 2X – 1 MUST BE GREATER
THAN OR EQUAL TO ZERO THIS RESTRICTION WILL HELP US
FIND OUR DOMAIN. WE JUST NEED TO SOLVE THIS
FOR X, SO WE’LL ADD 1 TO BOTH SIDES,
DIVIDE BY 2, SO X HAS TO BE GREATER THAN
OR EQUAL TO 1/2 WHICH WOULD BE DOMAIN OF F OF X. IF WE WANTED TO USE
INTERVAL NOTATION THIS WOULD BE THE INTERVAL FROM
1/2 TO INFINITY CLOSED ON 1/2, MEANING IT INCLUDES 1/2. TO FIND THE RANGE OF F OF X WE WANT TO DETERMINE
ALL THE POSSIBLE Y VALUES OF THIS FUNCTION. WELL THE VALUE OF THIS SQUARE
ROOT IS ALWAYS GOING TO BE GREATER THAN OR EQUAL TO ZERO SO THE SMALLEST VALUE
THIS COULD BE WOULD BE 0 – 3 WHICH WOULD BE -3 AND ALL OTHER FUNCTION VALUES
WOULD BE LARGER THAN -3. SO THE RANGE IS Y
IS GREATER THAN OR EQUAL TO -3 OR USING INTERVAL NOTATION WE WOULD HAVE THE INTERVAL
FROM -3 TO INFINITY. LET’S GO AHEAD AND VERIFY THIS BY GRAPHING
THE ORIGINAL FUNCTION. AGAIN,
HERE’S THE GIVEN FUNCTION, IF WE WERE TO PROJECT THIS
ON TO THE X AXIS NOTICE HOW THE X VALUES WOULD BE
FROM -1/2 TO INFINITY AND IF WE PROJECTED THIS
ONTO THE Y AXIS NOTICE HOW THE VALUES WOULD BE
FROM -3 TO POSITIVE INFINITY. SO WE HAVE THE DOMAIN AND RANGE
CORRECT, NOW LET’S GO AHEAD AND FIND
THE INVERSE FUNCTION. REMEMBER INVERSE FUNCTIONS
UNDO EACH OTHER, MEANING IF FUNCTION F HAS AN
INPUT OF X AND AN OUTPUT OF Y THIS Y BECOMES THE INPUT
INTO THE INVERSE FUNCTION WHICH RETURNS THE ORIGINAL VALUE
OF X. SO THE PROCESS FOR FINDING
THE INVERSE FUNCTION IS TO WRITE THIS
IN TERMS OF X AND Y AND THEN INTERCHANGE
THE X AND Y VARIABLES. SO WE CAN WRITE
THE GIVEN FUNCTION AS Y=THE SQUARE ROOT
OF THE QUANTITY (2X – 1) – 3 WHICH MEANS THE INVERSE FUNCTION
WOULD BE THE EQUATION X=THE SQUARE ROOT OF (2Y – 1) – 3. AND NOW WE JUST NEED TO SOLVE
THIS FOR Y AND REPLACE Y WITH
INVERSE FUNCTION NOTATION. SO THE FIRST STEP, WE’LL ADD 3
TO BOTH SIDES OF THE EQUATION. SO WE’LL HAVE X + 3=THE SQUARE
ROOT OF THE QUANTITY 2Y – 1. NOW TO UNDO THE SQUARE ROOT WE’LL SQUARE BOTH SIDES
OF THE EQUATION. LET’S GO AHEAD AND LEAVE THIS
AS THE QUANTITY X + 3 SQUARED. ON THE RIGHT SIDE,
SQUARING THE SQUARE ROOT LEAVES US WITH THE RADICAND
OF 2Y – 1. NEXT STEP, WE’LL ADD 1
TO BOTH SIDES OF THE EQUATION. IT’LL GIVE US THE QUANTITY
(X + 3 SQUARED) + 1=2Y. THE LAST STEP,
INSTEAD OF DIVIDING BY 2 LET’S MULTIPLY BOTH SIDES
BY 1/2. SO ON THE LEFT SIDE WE’D HAVE 1/2 x THE QUANTITY
X + 3 SQUARED + 1 EQUALS– ON THE RIGHT SIDE,
WE JUST HAVE Y. NOW WE COULD MULTIPLY THIS OUT
AND TRY TO SIMPLIFY THIS BUT I’M GOING TO GO AHEAD
AND LEAVE IT IN THIS FORM BUT WE SHOULD REPLACE Y
WITH INVERSE FUNCTION NOTATION. SO F INVERSE OF X IS EQUAL TO 1/2 x THE QUANTITY X
+ 3 SQUARED + 1. NOW IT’S IMPORTANT TO REMEMBER THE DOMAIN OF THIS INVERSE
FUNCTION IS RESTRICTED. IT’S GOING TO BE THE RANGE
OF THE ORIGINAL FUNCTION. SO LET’S GO AHEAD AND LIST THAT. THE DOMAIN OF THIS FUNCTION
IS GOING TO BE WHEN X IS GREATER THAN
OR EQUAL TO -3 OR THE INTERVAL FROM -3
TO INFINITY. AGAIN REMEMBER THE OUTPUT
OF THE ORIGINAL FUNCTION BECOMES THE INPUT
INTO THE INVERSE FUNCTION. SO THE RANGE OF F
IS THE DOMAIN OF F INVERSE AND THE RANGE
OF THE INVERSE FUNCTION WILL BE THE DOMAIN
OF THE ORIGINAL FUNCTION. SO OUR DOMAIN IS GOING TO BE Y
GREATER THAN OR EQUAL TO 1/2. WHICH USING INTERVAL NOTATION
WOULD BE FROM 1/2 CLOSED ON 1/2 TO POSITIVE INFINITY. SO HERE’S THE DOMAIN
OF THE ORIGINAL FUNCTION, HERE’S OUR INVERSE FUNCTION AND THE DOMAIN AND RANGE
OF THE INVERSE FUNCTION. THE LAST STEP, I ALWAYS LIKE
TO VERIFY THIS GRAPHICALLY. REMEMBER IF WE GRAPH
THE FUNCTION AND IT’S INVERSE FUNCTION
ON THE SAME COORDINATE PLAN THE TWO FUNCTIONS
SHOULD BE SYMMETRICAL ACROSS THE LINE Y=X. HERE’S THE GRAPH
OF THE ORIGINAL FUNCTION, OUR SQUARE ROOT FUNCTION AND HERE’S THE GRAPH
OF THE OUR QUADRATIC FUNCTION OR THE INVERSE FUNCTION AND NOTICE HOW THE DOMAIN
OF THE INVERSE FUNCTION HAS BEEN RESTRICTED DUE TO THE
RANGE OF THE ORIGINAL FUNCTION. AND HERE’S THE LINE Y=X AND NOTICE HOW THE TWO GRAPHS
ARE SYMMETRICAL ACROSS THIS LINE. OKAY, I HOPE YOU FOUND THIS

• ### Dr Philosophous

the domain was actually x is GREATER than 1/2 not greater than or equal too

• ### Jack Stone

Thank you very much !

• ### Absurd Usernam3

No, the guy is right. x can be greater than or equal to 1/2.

• ### nick ramos

THANK YOU! YOU REALLY HELPED ME ON MY HOMEWORK=)

• ### Tyra Holland

you have no idea how much you just helped me. thank you so much!

• ### A - B.O.M.B

I'm in Grade 11, and the graph at 1:45 is wrong. Domain is from 1/2 to infinity. 1/2 being the minimum value X could be

• ### Jed Gooridge

thanks but one question, do you have always have to multiple the Y by the Domain??

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