Question

4,Given the function

Answer 1

what?

Answer 2

\[f:x \rightarrow \frac{ 2x }{ x-1 }\],\[x \neq1\]
b)state the value of x such that f^-1 does not exist.

Answer 3

@jim_thompson5910

Answer 4

@sammixboo

Answer 5

well,first find f inverse

Answer 6

u meant this?yea?

Answer 7

the question said that f^-1 is not exist @PFEH.1999

Answer 8

The format of your question does not make sense.

\(\color{blue}{\text{Originally Posted by}}\) @MARC_
\[f:x \rightarrow \frac{ 2x }{ x-1 }\],\[x \neq1\]
b)state the value of x such that f^-1 does not exist.
\(\color{blue}{\text{End of Quote}}\)

That gets a little confusing to understand :\

Answer 9

yes @Jhannybean , i can't catch what is the question too ;)

Answer 10

the answer in the book said it's 2

Answer 11

but i keep getting =2/0

Answer 12

Is it \(f(x) = \dfrac{2x}{x-1}\) ?

Answer 13

yes ! now i got what is the question (maybe)
\[f ^{-1} (x)= \frac{ x }{ x-2 }\]

Answer 14

so for x=2 does not exist? u meant this?

Answer 15

it's \[f:x \rightarrow \frac{ 2x }{ x-1 }\]

Answer 16

what does the \(:\) and \(\rightarrow\) imply?

Answer 17

\[f(x)=\frac{ 2x }{ x-1 }\]
is the same as
\[f:x \rightarrow \frac{ 2x }{ x-1 }\]
@Jhannybean

Answer 18

@ganeshie8

Answer 19

maybe start by finding the inverse of f(x)

Answer 20

\[\large y = \dfrac{2x}{x-1}\]

remember how to find the inverse ?

Answer 21

you find it in two steps :
1) swap x and y
2) solve y

Answer 22

i got \[f^-1(x)=\frac{x }{ x-2}\]

Answer 23

correct ;)

Answer 24

Yes! look at the denominator, what are the excluded values of \(f^{-1}(x)\)

Answer 25

what value of \(x\) makes the denominator 0 ?

Answer 26

x=2

Answer 27

yes why is it a bad value ?

Answer 28

\(2\) is a bad input for \(f^{-1}(x)\) because it makes the denominator \(0\) and the function goes crazy when denominator approaches 0

Answer 29

maybe plugin a close number like x = 1.999 and see what the function outputs

Answer 30

\[\large f^{-1}(x)=\frac{x }{ x-2}\]

plugin x=1.999
\[\large f^{-1}(1.999)=\frac{1.999 }{ 1.999-2} =\frac{1.999 }{-0.001} = -1999 \]

Answer 31

plugin x = 1.99999

\[\large f^{-1}(1.99999)=\frac{1.99999 }{ 1.99999-2} =\frac{1.999 }{-0.00001} = -199999 \]

Answer 32

as you can see the closer you get to 2, the more crazy huge negative value the function is taking so we say x = 2 is an exclude value for the function

Answer 33

graph the function in your favorite desmos and see whats going on at x = 2

Answer 34

okay

Answer 35

in calculas we say when x approaches 2 then f^-1 (x) approaches to \( \infty\)! see the graph:

Answer 36

* using https://www.desmos.com/calculator

Answer 37

it's a straight line

Answer 38

which mean it's undefined right?

Answer 39

yes , at x=2 then f^-1 (x) is undefined.

Answer 40

\(\bf\huge\color{#ff0000}{T}\color{#ff2000}{h}\color{#ff4000}{a}\color{#ff5f00}{n}\color{#ff7f00}{k}~\color{#ffaa00}{Y}\color{#ffd400}{o}\color{#bfff00}{u}\color{#4600ff}{!}\color{#6800ff}{!}\color{#8b00ff}{!}\) @ganeshie8 @PFEH.1999 @Jhannybean