C3 Functions

Question part a):

Answer

Spoiler

Alright,

f^{-1}(5) = 2

and as 2 is an element from the range of

f^{-1}

meaning that it's an element from the domain from

f(x)

(as the range of

f^{-1} =

the domain of

f(x)

) but since

f(x)

has the restriction

x < 0

this means that the inverse must be the negative square root?

How would you figure out whether to take the positive or negative square root as the inverse though? Would it be by recognition? Is there another way besides recognition? Would you sub in x = 0 or any specific values to check?

(edited 10 years ago)

Reply 1

electriic_ink

15

Because if it wasn't,

f^{-1}(5)=2

. So f(2)=5. But f is not defined when x=2 (it's only defined when x<0).

Reply 2

singapaul

8

Because the domain states that X<0 so it has to be the negative square root, or else the final answer would be greater than 0

Reply 3

Noah~

OP

Original post by electriic_ink

Because if it wasn't,

f^{-1}(5)=2

. So f(2)=5. But f is not defined when x=2 (it's only defined when x<0).

I don't quite fully understand. You randomly put the input as 5, what if the input was a set of different numbers?

Original post by singapaul

Because the domain states that X<0 so it has to be the negative square root, or else the final answer would be greater than 0

Could you explain further (I still do not quite fully get the gist of this)

Reply 4

Noah~

OP

Original post by electriic_ink

Because if it wasn't,

f^{-1}(5)=2

. So f(2)=5. But f is not defined when x=2 (it's only defined when x<0).

Original post by singapaul

Because the domain states that X<0 so it has to be the negative square root, or else the final answer would be greater than 0

Alright,

f^{-1}(5) = 2

and as 2 is an element from the range of

f^{-1}

meaning that it's an element from the domain from

f(x)

(as the range of

f^{-1} =

the domain of

f(x)

) but since

f(x)

has the restriction

x < 0

this means that the inverse must be the negative square root?

How would you figure out whether to take the positive or negative square root as the inverse though? Would it be by recognition? Is there another way besides recognition? Would you sub in x = 0 or any specific values to check?

(edited 10 years ago)

Reply 5

electriic_ink

15

Original post by Noah~

Alright,

\mathbf{f^{-1}(5) = 2}

and as 2 is an element from the range of

f^{-1}

meaning that it's an element from the domain from

f(x)

(as the range of

f^{-1} =

the domain of

f(x)

) but since

f(x)

has the restriction

x < 0

this means that the inverse must be the negative square root?

No, f^{-1}(5)=-2.

If you don't know what f^{-1} is, there are two options for f^{-1}(5) -- +2 if you take the positive square root and -2 if you take the negative. You know that f^{-1}(5)=-2 (as -2 is in the domain of f but 2 isn't) and therefore you have to take the negative square root.

How would you figure out whether to take the positive or negative square root as the inverse though? Would it be by recognition? Is there another way besides recognition? Would you sub in x = 0 or any specific values to check?

The "proper" argument (that will work regardless of the function) is that both

f\circ f^{-1}(x)=x

and

f^{-1} \circ f(x)=x

. If you picked the positive square root, when you try to work out

f\circ f^{-1}(x)

, you find that

f^{-1}(x)>0

and so

f\circ f^{-1}(x)

is never defined (and therefore cannot be x).

So you need to consider each possibility and, by looking at the ranges/domains, deduce for which one

f\circ f^{-1}(x)=x

and

f^{-1} \circ f(x)=x

hold.

But in practice, I think you can get away with trying out a few random values.

(edited 10 years ago)

C3 Functions

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you