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Let \(f(x) = \frac{2x + 3}{kx - 2}\). Find all real numbers \(k\) so that \(f^{-1}(x) = f(x)\).

 Feb 18, 2019
 #1
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To find the inverse of f(x), set f(x) = y. Then solve for x in terms of y. 

 

By doing this I got \(x=\frac{2y+3}{ky-2}\), so \(f^{-1}(x)=\frac{2x+3}{kx-2}\).

 

This means that \(k\) can be any real number for \(f(x) = f^{-1}(x)\) (as long as \(x\ne\frac{2}{k}\)).

 Feb 18, 2019
 #2
avatar+104836 
+2

We can write

 

y = [ 2x + 3 ] / [ kx - 2]        isolate x

 

y [ kx - 2 ] =  2x + 3

 

(ky)x - 2y = 2x + 3

 

(ky)x - 2x =  2y + 3

 

(ky - 2) x =  2y + 3    

 

x =  (2y + 3) / ( ky - 2)    "swap" x and y

 

y = (2x + 3) / (kx - 2)  =  the inverse

 

The inverse is the same as f(x)

 

So.....k can take on any real value

 

 

 

cool cool cool

 Feb 18, 2019
 #3
avatar+23313 
+4

help
Let

\(\large{f(x) = \dfrac{2x + 3}{kx - 2}}\).

Find all real numbers

\(\mathbf{k}\)

so that

\(\large{f^{-1}(x) = f(x)}\).

 

discontinuity:

\(\begin{array}{|rcll|} \hline 2x+3 &=& 0 \\ 2x &=& -3 \\ x &=& -\dfrac{3}{2} \\ \hline kx - 2 &\ne& 0 \\ kx &\ne& 2 \\ k &\ne& \dfrac{2}{x} \quad | \quad x = -\dfrac{3}{2} \\\\ k &\ne& \dfrac{2}{-\dfrac{3}{2}} \\\\ \mathbf{k} & \mathbf{\ne} & \mathbf{-\dfrac{4}{3}} \\ \hline \end{array} \)

 

At \(k = -\dfrac{4}{3} \Rightarrow x = -\dfrac{3}{2} \Rightarrow y = \dfrac{0}{0}\)

 

laugh

 Feb 19, 2019

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