Thread: Two Equations (solving for inverses)

1. Two Equations (solving for inverses)

Find the inverse equations of the following:

1. $f(x) = 3x + ln(x)$

2. $g(x) = \frac{x+1}{2x+1}$

So the equations become...

1. $x = 3y + ln(y)$

2. $x = \frac{y + 1}{2y + 1}$

Sovle for $y$.

I've played with the algebra a few different ways, especially on the rational function, but I still can't solve for 'y' in either equation! I've also graphed both functions, and they both appear to have an inverse (geometrically anyway).

Step-by-step algebra would be helpful - thanks!

2. [quote=TaylorM0192;446936]Find the inverse equations of the following:

1. $f(x) = 3x + ln(x)$

2. $g(x) = \frac{x+1}{2x+1}$

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2.

$g^{-1}(x) = \frac{1-x}{2 x-1}$

Originally Posted by TaylorM0192
Find the inverse equations of the following:

1. $f(x) = 3x + ln(x)$

2. $g(x) = \frac{x+1}{2x+1}$

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1.

$f^{-1}(x) = 1/3 W(3 e^x)$

2.

$g^{-1}(x) = \frac{1-x}{2 x-1}$
Your $g^{-1}(x)=\frac{1-x}{2x-1}$ cannot be the inverse of my $g(x)=\frac{x+1}{2x+1}$ (confirmed by the graphs of each & algebraically by $g(g^-1(x))$ not being equal to 'x').

I don't know what $f^{-1}(x) = 1/3 W(3 e^x)$ is...

4. Figured out $g^-1(x) = \frac{x-1}{1-2x}$.

Still don't know $f^-1(x)$ though.

5. Originally Posted by TaylorM0192
Figured out $g^-1(x) = \frac{x-1}{1-2x}$.

Still don't know $f^-1(x)$ though.
$f^{-1}(x) = \frac {1}{3} W(3 e^x)$

W is the product log function