– 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

HELPFUL.

## 10 Comments

## 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??

## Nissy Radovic

thank you very helpful!

## Melissa Santa Cruz

how would you solve this sqrt[5]{n-3}-2 ( translates to g(n) = square root of n – 3 with an index of five minus 2 ( minus 2 is not under the radical btw)

## _ tomato _

This video was super helpful, only thing that actually did