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# help

+1
51
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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
+98126
+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

Feb 18, 2019
#3
+21848
+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 $$k = -\dfrac{4}{3} \Rightarrow x = -\dfrac{3}{2} \Rightarrow y = \dfrac{0}{0}$$

Feb 19, 2019