anonymous
  • 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.
Mathematics
schrodinger
  • schrodinger
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anonymous
  • 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
  • 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
  • anonymous
y=f(x) is invertable if and only if f(x) is one-one and onto

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