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- anonymous

True or False:
For a trigonometric function, y = f(x), then x = F^-1(y). Explain your answer. (the capital F denotes function, not relation)
For a one-to-one function, y = f(x), then x = f^-1(y). Explain your answer.
For any function, x = f^-1(y), then y = f(x). Explain your answer.

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- anonymous

- schrodinger

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- anonymous

for a function
\[y=f(x)\]
we can say
\[x=f^{-1}(y)\]
\[\iff\](this symbol means if and only if, it's nice to remember some mathematical symbols on the go!)
\[f(x)\]
is one-to-one AND onto
both conditions must be satisfied simultaneously
Now what this means?
A function is said to be one-to-one(or one-one), if a number in the range of the function is only linked to a specific number in the domain
Think of it this way, if you were to find inverse of a number from the range of f(x), if it were linked to multiple numbers, then it would not be an inverse function, because a function by definition takes an input and gives only 1 output
consider the drawing:
|dw:1439876556798:dw|
Now example of an one-one function:
|dw:1439876737122:dw|
Onto is when EACH number in the range has a link to a number in domain
Both drawing 1 and 2 above are not onto because there are elements which have no links back to set x, (eg. v)
Think of it like this in drawing 2
if you were to calculate the inverse at v, it would not be defined as v is not linked to some number in set x, thus your inverse function is not defined, so a function must be onto
Eg.
|dw:1439877071440:dw|
Think about these points and try to attempt the question

- anonymous

So y=f(x) can only be inverted to x=f^-1 (y) when f(x) is one to one? Here's what I have so far, but I'm very unsure about it.
1. y = f(x), then x = F^-1(y) is true because capital F requires the equation to be a one to one function, not a relation.
2. y = f(x), then x = f^-1(y). True, because it's stated in the question that it is a one to one function and thus can be inverted.
3. x = f^-1(y), then y = f(x). False, because we do not know whether it is a function or relation and cannot determine if it can be inverted.

- anonymous

y=f(x) is invertable if and only if f(x) is one-one and onto

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- anonymous

example of a function whose inverse can exist:
|dw:1439879124917:dw|

- anonymous

for example consider the sine function
\[y=\sin(x)\]
At x=pi/4 and 3pi/4 we have
\[y=\sin(\frac{\pi}{4})=\frac{1}{\sqrt{2}}\]
\[y=\sin(\frac{3\pi}{4})=\frac{1}{\sqrt{2}}\]
Now this shows that the sine function is not one-one,
if we want to evaluate
\[f^{-1}(\frac{1}{\sqrt{2}})\]
we are getting 2 answers pi/4 and 3pi/4, but a function only gives 1 answer
therefore for inverse trigonometric functions, we have to apply certain restrictions about which you would study later if you've already not studied it

- anonymous

thus for your question 1)
trig functions are not one-one so that already tells you their inverse cannot exist unless we apply some restrictions
so it is FALSE
For your question 2)
A function must be both one-one and onto, since the question says function is one one that's 1 condition satisfied, since it's not given that the function is onto, we assume it to be non-onto therefore inverse does not exist
so FALSE
for your question 3)
False, not every function has an inverse, some functions are not one-one, some are not onto and some are neither.

- anonymous

Ah, my textbook only covered the definition of one-to-one, it didn't even mention onto! Thank you so much for explaining it! So we basically assume unless stated that a function fulfills both conditions, it cannot be inverse? I hope I got it now!

- anonymous

Yes, a function must satisfy both conditions

- anonymous

of one-one and onto

- anonymous

I hope you understood the sine example, the reason why trig functions are not one-one

- anonymous

Since your textbook only covers one-one, you may consider Q2 to be as true actually(as per your textbook)

- anonymous

A function that is one-one is also referred to as injective function
and a function that is onto is also referred to as surjective function
So inverse exists for a function that is both injective and surjective
you can read more about it here:
http://www.regentsprep.org/regents/math/algtrig/atp5/OntoFunctions.htm

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