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.
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
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.
y=f(x) is invertable if and only if f(x) is one-one and onto
example of a function whose inverse can exist: |dw:1439879124917:dw|
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
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.
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!
Yes, a function must satisfy both conditions
of one-one and onto
I hope you understood the sine example, the reason why trig functions are not one-one
Since your textbook only covers one-one, you may consider Q2 to be as true actually(as per your textbook)
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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