- #1
ColdheartedGod
- 5
- 0
- Homework Statement
- Find the inverse function for a given function.
- Relevant Equations
- f(x) =(x+(x^2+1)) ^(1/3)+(x-(x^2+1))^(1/3)
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Did you try to write out ##y^3## ?ColdheartedGod said:other approach
AndColdheartedGod said:I'm so sorry
Example: (right-click on formula to show math as ##\TeX## comands)BvU said:1. Learn to post using ##\LaTeX##
Thanks for the explanations!BvU said:Concerning
And
Example: (right-click on formula to show math as ##\TeX## comands)
$$y = \sqrt[\leftroot{-3}\uproot{12}3]{x+\sqrt{x^2+1}} + \sqrt[\leftroot{-3}\uproot{12}3]{x-\sqrt{x^2+1}} +$$
Or simpler: your f(x) =(x+(x^2+1)) ^{(1/3)}+(x-(x^2+1))^{(1/3)} -- adorned with {} to group the (1/3) and enclosed in double $$ becomes
$$f(x) =(x+(x^2+1)) ^{1/3}+(x-(x^2+1))^{1/3} $$
when it should have been
$$f(x) =\left(x+(x^2+1)^{1/2}\right ) ^{1/3}+\left(x-(x^2+1)^{1/2}\right)^{1/3} $$
Enough showing off; back to the exercise at hand
Any light at the end of the tunnel ?
ColdheartedGod said:Thanks for the explanations!
Well, it seems that my current knowledge in trigonometry isn't good enough to carry that part so I'm now struggling to find out more about sinh functions and how to apply it in this case.
It is worth trying!epenguin said:Not that anything is guaranteed before you try, but have you tried straight algebra?
In order to not get over-long formulae I would write this as
y = (A⅓ + B⅓) at first, then look at y3. (Even if, as I see, this has been disadvised.)
I'll third this recommendation. I didn't work it out, but I think I see how it goes. You get some nice simplification.epenguin said:Not that anything is guaranteed before you try, but have you tried straight algebra?
Not to mention that an inverse does not necessarily exist, in the reals, for all y. However to see into that in this case I think you need to make progress in the elaboration along the lines indicated.epenguin said:Note that inverses are not necessarily unique, but if you even get one that would be something.
epenguin said:Not to mention that an inverse does not necessarily exist, in the reals, for all y. However to see into that in this case I think you need to make progress in the elaboration along the lines indicated.
You can easily see the function f(x) itself does not exist in the reals for x between -1 and 1.
vela said:I'll third this recommendation. I didn't work it out, but I think I see how it goes. You get some nice simplification.
epenguin said:You can easily see the function f(x) itself does not exist in the reals for x between -1 and 1.
PeroK said:##f(0) = 0##
What's the problem?
It is not a good idea. Just take the cube of both sides as they are. The cube of ##y=a+b## is ##y^3=a^3+b^3+3a^2b+3ab^2 ## which can be written as ##y^3=a^3+b^3+3(ab)(a+b)##. By substitution, it becomes very simple and x can be easily isolated.WWGD said:Maybe you can follow recommendation of cubing by first leaving one of the radical expressions alone in one side before doing the cubing to avoid a mess of an expression.
You indeed get massive cancellation, but there's still a cube root left over. I get a cubic in x, lightly sprinkled with powers of y^3. I could now use cardano's formula, and maybe there's even more cancellation?ehild said:It is not a good idea. Just take the cube of both sides as they are. The cube of ##y=a+b## is ##y^3=a^3+b^3+3a^2b+3ab^2 ## which can be written as ##y^3=a^3+b^3+3(ab)(a+b)##. By substitution, it becomes very simple and x can be easily isolated.
Haven't seen much follow-up for this one. even though ##\sinh + \cosh## and ##\sinh - \cosh## yield such nice expressions ...BvU said:Other tack: try ##x = \cosh(z)##
[edit] I mean the other one: sinh !
I solved using that substitution, got my answer deleted though, as per the rulesBvU said:Haven't seen much follow-up for this one. even though ##\sinh + \cosh## and ##\sinh - \cosh## yield such nice expressions ...
I saw it, but it was a bit complicated. The solution is quite simple just a polynomial of y.archaic said:I solved using that substitution, got my answer deleted though, as per the rules
No need for anything complicated, like Cardano formula. If you follow the hint in #Post 20, the only cube roots remaining are in a+b, which is identical with y.willem2 said:You indeed get massive cancellation, but there's still a cube root left over. I get a cubic in x, lightly sprinkled with powers of y^3. I could now use cardano's formula, and maybe there's even more cancellation?
There is however an easier way. If you look at the original expression, you can see that it looks just like the outcome of cardano's formula. There is a depressed cubic in y with x as one of the parameters, which will have your expression as the real root. Solving this for x will be trivial.
To determine if a function has an inverse, you can use the horizontal line test. If a horizontal line intersects the graph of the function at more than one point, then the function does not have an inverse. If the horizontal line only intersects the graph at one point, then the function does have an inverse.
The process for finding the inverse of a function involves swapping the x and y variables and solving for y. This will result in a new equation, which represents the inverse of the original function.
No, not all functions can be inverted. In order for a function to have an inverse, it must pass the horizontal line test and be one-to-one, meaning that each input has a unique output.
Some common methods for finding the inverse of a function include using algebraic manipulation, graphing, and using the inverse function theorem. In more complicated cases, numerical methods may also be used.
Yes, there are special techniques such as using substitution, partial fractions, and trigonometric identities to simplify the function before finding the inverse. Additionally, some functions may require the use of logarithms or exponential functions in order to find the inverse.