# Thread: Value of m for a function to be its own inverse.

1. ## Value of m for a function to be its own inverse.

Given,

y = f(x) = (3x - 5) / (2x - m)

Find the value of m such that f(y) = x

I tried doing it myself, but it just kept getting lengthier & didn't seem to end..

Any idea how to solve that?

2. What exactly did you try?

3. If f(x) = (3x - 5) / (2x - m), you should know what f(y) is. Equate that with x; and find m. This is not at all a lengthy procedure.

4. This is what I did: -

If y = f(x) = (3x - 5) / (2x - m) ----- (1)

then, f(y) = (3y - 5) / (2y - m) = x [given condition]

or, [3 {(3x - 5) / (2x - m)} - 5] / [2 {(3x - 5) / (2x - m)}] = x [putting the value of y from eqn. (1)]

then....

this is where I came: -

(-x) + 5m - 15 = (6x^2) - 10x - (2x^2)m + (m^2)x

This is where I got stuck & couldn't farther anymore steps..

1. You should have

$\displaystyle \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x,$ not

$\displaystyle \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}}=x.$

2. Once you carry this correction through, you can equate like powers of x to get the value for $\displaystyle m.$

6. Originally Posted by Ackbeet

1. You should have

$\displaystyle \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x,$ not

$\displaystyle \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}}=x.$

2. Once you carry this correction through, you can equate like powers of x to get the value for $\displaystyle m.$

Thanks for the correction.

I did do $\displaystyle \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x,$, and not $\displaystyle \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}}=x.$

Because only after doing $\displaystyle \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x,$ , can you arrive at (-x) + 5m - 15 = (6x^2) - 10x - (2x^2)m + (m^2)x.

It's just that I forgot to write it here as I'm new to this web.

However, could you just give a simple example to show how to equate the powers of X on both sides of an equation?

I know I learned that at Grade 8 or 9.. But I don't remember the exact procedure.

So if you could just give a simple example of it, that would be very helpful..

7. Originally Posted by arijit2005
Thanks for the correction.

I did do $\displaystyle \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x,$, and not $\displaystyle \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}}=x.$

Because only after doing $\displaystyle \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x,$ , can you arrive at (-x) + 5m - 15 = (6x^2) - 10x - (2x^2)m + (m^2)x.

It's just that I forgot to write it here as I'm new to this web.

However, could you just give a simple example to show how to equate the powers of X on both sides of an equation?

I know I learned that at Grade 8 or 9.. But I don't remember the exact procedure.

So if you could just give a simple example of it, that would be very helpful..

No problem. In your equation, which looks correct, you can scan the whole equation for the power x^2, and equate all those powers. On the LHS, you don't have any. On the RHS, you have 6-2m. Therefore, 0 = 6-2m.

Next, you can equate the single powers of x^1. On the LHS, you have -1, and on the RHS, you have -10+m^2. Hence, you have the equation -1 = -10 + m^2.

Finally, you can equate the constants (powers of x^0). On the LHS, you have 5m-15, and on the RHS, you have 0. Hence, 5m-15 = 0.

Thankfully, or perhaps the problem was designed with this in view, there is only one value of m that satisfies all three equations, but it does satisfy all three equations. You'd certainly never expect there to be a solution, since over-determined systems of equations often don't have solutions. However, this system does.

Make sense?

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### y=f(x) = (3x-5)/(2x-m) and x= f(y) then find m

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