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If $$f(x)=\dfrac{a}{x+2}$$, solve for the value of a so that $$f(0)=f^{-1}(3a)$$.

Feb 18, 2019

#1
+6187
+1

$$\text{Well the brute force way is }\\ y=\dfrac{a}{x+2}\\ \dfrac 1 y = \dfrac{x+2}{a}\\ x = \dfrac a y - 2\\ f^{-1}(x) = \dfrac{a}{x} - 2$$

$$f(0) = f^{-1}(3a)\\ \dfrac{a}{2} = \dfrac{a}{3a}-2\\ \dfrac a 2 = \dfrac 1 3 - 2 = -\dfrac 5 3\\ a = - \dfrac{10}{3}$$

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Feb 18, 2019
#2
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If  $$f(0)=f^{-1}(3a)$$  ,  solve for the value of a so that    $$f(0)=f^{-1}(3a)$$

This looks kinda interesting   -  I will assume that a is a constant.

$$let\;\\ y=\dfrac{a}{x+2} \quad x \ne-2\\ x+2=\dfrac{a}{y}\\ x=\dfrac{a}{y}-2\\ f^{-1}(x)=\dfrac{a}{x}-2\\~\\ f^{-1}(3a)=\dfrac{a}{3a}-2=\frac{1}{3}-2=\frac{-5}{3}\\$$

$$f(0)=\frac{a}{2}$$

if

$$f(0)=f^{-1}(3a)\\ \frac{a}{2}=\frac{-5}{3}\\ a=\frac{-5*2}{3}\\ a=\frac{-10}{3}\\ a=-3\frac{1}{3}$$

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Feb 18, 2019
edited by Melody  Feb 18, 2019
#3
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You beat me Rom

Feb 18, 2019
#4
+6187
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Nah I'd never beat you.  Taunt you relentlessly perhaps but never beat you!

Rom  Feb 18, 2019
#5
+109524
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LOL you have never had to put up with me full time :)

Melody  Feb 18, 2019
#6
+6187
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btw there's a small error in your answer.. I'll let you find it!

Rom  Feb 18, 2019
#7
+109524
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Thanks Rom, I fixed it.

My own little errors (or even big ones) don't usually trouble me that much.

I always think it is the asker's job to decide how much of the answer they should accept.

But I do not mind if people correct me or let me know. Thank you

Melody  Feb 18, 2019
#8
+24983
+3

If

$$\large{ f(x)=\dfrac{a}{x+2}}$$,

solve for the value of $$\mathbf{a}$$ so that

$$\large{f(0)=f^{-1}(3a)}$$.

$$\begin{array}{|rcll|} \hline f\Big( f^{-1}(x) \Big) &=& x \quad | \quad x = 3a \\ f\Big( f^{-1}(3a) \Big) &=& 3a \quad | \quad f^{-1}(3a)=f(0) \\ f\Big( f(0) \Big) &=& 3a \quad | \quad f(0) = \dfrac{a}{0+2} = \dfrac{a}{2} \\ f\left( \dfrac{a}{2} \right) &=& 3a \quad | \quad f\left(\dfrac{a}{2} \right) = \dfrac{a}{\dfrac{a}{2}+2} = \dfrac{2a}{4+a} \\ \dfrac{2a}{4+a} &=& 3a \quad | \quad :a \\\\ \dfrac{2 }{4+a} &=& 3 \\\\ 4+a &=& \dfrac{2}{3} \\\\ a &=& \dfrac{2}{3} -4 \\\\ a &=& \dfrac{2}{3} -\dfrac{12}{3} \\\\ \mathbf{ a } &\mathbf{=}&\mathbf{ -\dfrac{10}{3}} \\ \hline \end{array}$$

Feb 18, 2019