Its been awhile since i've done any math, (2 years). Can someone get me started in finding the inverse of this equation, I know you switch the x with y's, but i'm stuck from there.
Help me find inverse of x^3 + x (Need a clue)
- Thread starter Prince of Persia
- Start date
GTaudiophile
Lifer
- Oct 24, 2000
- 29,767
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I don't know if that's what you need in this case. But lately in math we've been finding reciprocals of certain fractions (in your case it's all over 1), etc. So if the 'inverse' is just like the reciprocal, or negative reciprocal, I don't know - try flipping it. See what results you get on the graph, anyway. Eventually you're bound to get the right answer. 
-RSI
-RSI
The way I learned for an inverse of the function is just switch the Xs with the Ys and then solve
Hope that helps!
Hope that helps!
<< The way I learned for an inverse of the function is just switch the Xs with the Ys and then solve
Hope that helps! >>
Ahhh, that's probably right. But I'd ask someone that's doing that right now. I'll probably be doing inverses of functions later on.
-RSI
<< The way I learned for an inverse of the function is just switch the Xs with the Ys and then solve
Hope that helps! >>
EXACTLY! that's how you do inverse functions
go do it NOW NOW NOWhere's an example if you dont' understand
x + 1 = y becomes y + 1 = x
okay bai bai me go now
but i know that i switch the x and y's and i'm left with
y^3 + y = x
but i can't figure out how to get x by itself.
y^3 + y = x
but i can't figure out how to get x by itself.
Umm.. I don't think that's the answer. An inverse of a function does the exact opposite of a function, i.e. inverse of x^2 is sqrt( x ) because:
y = x^2
substitute:
x = z^2
sqrt( x ) = z
z = inverse of y
So you do:
y = z ^ 2 = sqrt( x ) ^ 2 = y.
An inverse of a function, substituted into the function, results in an equality like y = y.
At least that's how I remember inverses... I may be wrong though. I couldn't find the inverse of x^3 + x though. Bah!
y = x^2
substitute:
x = z^2
sqrt( x ) = z
z = inverse of y
So you do:
y = z ^ 2 = sqrt( x ) ^ 2 = y.
An inverse of a function, substituted into the function, results in an equality like y = y.
At least that's how I remember inverses... I may be wrong though. I couldn't find the inverse of x^3 + x though. Bah!
This is how you do the inverse.
y=x^3+x
Swith the x and y.
x=y^3+y
Factor a y.
x=y(y^2+1)
Done.
Edit: This is from the above post and has nothing to do with the original problem.
For y=x^2,
x=y^2
y=sqrt(x) and -sqrt(x)
y=x^3+x
Swith the x and y.
x=y^3+y
Factor a y.
x=y(y^2+1)
Done.
Edit: This is from the above post and has nothing to do with the original problem.
For y=x^2,
x=y^2
y=sqrt(x) and -sqrt(x)
I respectfully disagree.
The problem was: find the inverse of y = x ^ 3 + x = x ( x ^ 2 + 1 )
Your answer is: x = y ( y ^ 2 + 1 )
So, you went on to do: inverse of y = x ^ 2
Your answer is: y = sqrt( x ) or y = - sqrt( x )
The problem is that in your first solution, you did not solve for y. In the second solution, you did. There seems to be a discrepancy.
The problem was: find the inverse of y = x ^ 3 + x = x ( x ^ 2 + 1 )
Your answer is: x = y ( y ^ 2 + 1 )
So, you went on to do: inverse of y = x ^ 2
Your answer is: y = sqrt( x ) or y = - sqrt( x )
The problem is that in your first solution, you did not solve for y. In the second solution, you did. There seems to be a discrepancy.
Errr...it can be solved--this is NOT a transcendental...buut...I'm not gonna claim to remember how
I cant remember howto isolate y...arrgg
I cant remember howto isolate y...arrgg
<< This is how you do the inverse.
y=x^3+x
Swith the x and y.
x=y^3+y
Factor a y.
x=y(y^2+1)
Done. >>
Done incorrectly heh.
<< Umm.. I don't think that's the answer. An inverse of a function does the exact opposite of a function, i.e. inverse of x^2 is sqrt( x ) because:
y = x^2
substitute:
x = z^2
sqrt( x ) = z
z = inverse of y
So you do:
y = z ^ 2 = sqrt( x ) ^ 2 = y.
An inverse of a function, substituted into the function, results in an equality like y = y.
At least that's how I remember inverses... I may be wrong though. I couldn't find the inverse of x^3 + x though. Bah! >>
An inverse of a function does not do the exact opposite of a function. If you graph y = x^2 and y = sqrt(x), they hardly look like opposites. I was always taught to find the inverse of a function by swapping the x and y variables (or whatever other variables are being used).
<< I respectfully disagree.
The problem was: find the inverse of y = x ^ 3 + x = x ( x ^ 2 + 1 )
Your answer is: x = y ( y ^ 2 + 1 )
So, you went on to do: inverse of y = x ^ 2
Your answer is: y = sqrt( x ) or y = - sqrt( x )
The problem is that in your first solution, you did not solve for y. In the second solution, you did. There seems to be a discrepancy. >>
You do not necessarily need to solve for y to find the inverse of a function. Either way the function is still the same. For example, say your original function is y = 2x. The inverse would be x = 2y. Even though I didn't solve for y in the inverse function, there is really no need to, because (x = 2y) is the same as (y = x / 2). The only difference is that in the first function, it is stated in terms of x. The second is stated in terms of y. They are still equal, therefore, it is irrelevant what variable is solved for.
<<
<< This is how you do the inverse.
y=x^3+x
Swith the x and y.
x=y^3+y
Factor a y.
x=y(y^2+1)
Done. >>
Done incorrectly heh. >>
So, then how do you do it, Mr. Smart?
<< Its been awhile since i've done any math, (2 years). Can someone get me started in finding the inverse of this equation, I know you switch the x with y's, but i'm stuck from there. >>
y = x^3 + x
I can't get past:
Y = (cube root)(x-y)
damn
the inverse is not a function (it should fail the vertical line test, but too lazy to graph). Therefore you will not be able to get y = blah*blah.
The inverse ought to be a function since the original function passes the horizontal line test.
it looks like:
|
|
|
/
/
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|
As i recall, the inverse of a function is a reflection over y=x, so shouldn't the inverse look like:
__________
\____________
if so, it passes the VLT, and therefore is a function.
m00se
it looks like:
|
|
|
/
/
|
|
|
As i recall, the inverse of a function is a reflection over y=x, so shouldn't the inverse look like:
__________
\____________
if so, it passes the VLT, and therefore is a function.
m00se
<< The inverse ought to be a function since the original function passes the horizontal line test.
it looks like:
|
|
|
/
/
|
|
|
As i recall, the inverse of a function is a reflection over y=x, so shouldn't the inverse look like:
__________
\____________
if so, it passes the VLT, and therefore is a function.
m00se >>
Yes, but, it isn't a point-to-point function, which means that the inverse will not be a function. However, that doesn't mean that there isn't an inverse, it's just that the inverse is not a function.
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