Question

PLEASE HELP WITH FOR MY FINAL.
how do i find the inverse of this?
f(x)=√x+1

Answer 1

\[f ^{-1}(x)=(x+1）^{2}\]

Answer 2

sorry, (x-1)^2

Answer 3

you take +1 to the other side and square to get rid of the sqrae rroot

Answer 4

where do i add +1? to the x?

Answer 5

depends if i tis \[f(x)=\sqrt{x+1}\] or \[f(x)=\sqrt{x}+1\]

Answer 6

its the first one

Answer 7

then just like solving say \[\sqrt{x+1}=7\]
square, then subtract 1
only this time solve \[\sqrt{y+1}=x\] for \(y\) in the same two steps

Answer 8

how did u get 7?

Answer 9

i made it up

Answer 10

can you solve it? can you solve \[\sqrt{x+1}=7\] for \(x\)?

Answer 11

you would subtract 1 from 7 right?

Answer 12

no not as a first step
you cannot subtract 1 because it is inside the radical, not outside

Answer 13

lets try this
we have \[f(x)=\sqrt{x+1}\] what is \(f(48)\)?

Answer 14

oh ok so x+1 is 2 right? and then √2 ?

Answer 15

um no, \(x+1=2\) only if \(x=1\) but \(x\) is a variable, it can be any number

Answer 16

for example, \(x\) could be \(48\)
what is \(f(48)\)?

Answer 17

you want me to square 48?

Answer 18

not for this function no
the function is add one, take the square root
for example if \[f(x)=\sqrt{x+1}\] then \[f(3)=\sqrt{3+1}=\sqrt4=2\]

Answer 19

and say \[f(8)=\sqrt{8+1}=\sqrt9=3\]

Answer 20

how about \(f(15)\)?

Answer 21

f(15) = √15+1 = √16 = 4

Answer 22

yay

Answer 23

how about \(f(48)\)?

Answer 24

f(x) = √x+1 = √1 = 1 right?

Answer 25

still asking what is \(f(48)\)

Answer 26

f(48) = √48 + 1 = √49 = 7

Answer 27

yes, good
so now we know if \(x=48\) then \(f(48)=7\)

Answer 28

now we have to work backwards
suppose you have to solve \[\sqrt{x+1}=7\] for \(x\) (we already know the answer is 48)
how would we find \(x\) in two steps?

Answer 29

subtract 1 from 7?

Answer 30

that give six, that is not going to help finding the 48 is it?

Answer 31

the last step you have was \(\sqrt{49}=7\) above
how can you get from 7 to 49?

Answer 32

7 times 7?

Answer 33

right, aka \(7^2\)

Answer 34

now how can you get from 49 to 48?

Answer 35

subtract by 1

Answer 36

exaclty
steps were
square, subtract 1
same steps here \[\sqrt{y+1}=x\]solve for \(y\) using the same two steps

Answer 37

so you do x^2?

Answer 38

yes that is first, that gives
\[y+1=x^2\] next step?

Answer 39

and then subtract 1 by x^2

Answer 40

well, i would say "subtract one FROM \(x^2\) but that is probably what you meant
how to you write it?

Answer 41

yeah sry

Answer 42

ok so what do you get as an answer?\[y=?\]

Answer 43

1?

Answer 44

no, not a number, how do you write "square x, subtract 1"

Answer 45

ohh ok. x2-1

Answer 46

x^2*

Answer 47

yay

Answer 48

that is your answer
if \[f(x)=\sqrt{x+1}\] then \[f^{-1}(x)=x^2-1\]

Answer 49

thanks so much! i get it now :D

Answer 50

your welcome~!
good luck with the next one

Answer 51

@satellite73 ccould u help me on this too? f(x) = x^2/2