Let \(f(x) = \frac{2x + 3}{kx - 2}\). Find all real numbers \(k\) so that \(f^{-1}(x) = f(x)\).
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 \(x\ne\frac{2}{k}\)).
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
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 \(k = -\dfrac{4}{3} \Rightarrow x = -\dfrac{3}{2} \Rightarrow y = \dfrac{0}{0}\)