We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
179
8
avatar

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

 Feb 18, 2019
 #1
avatar+6045 
+1

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

 

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

.
 Feb 18, 2019
 #2
avatar+105468 
0

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

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

.
 Feb 18, 2019
edited by Melody  Feb 18, 2019
 #3
avatar+105468 
-1

You beat me Rom   laugh

 Feb 18, 2019
 #4
avatar+6045 
+1

Nah I'd never beat you.  Taunt you relentlessly perhaps but never beat you!

Rom  Feb 18, 2019
 #5
avatar+105468 
-1

LOL you have never had to put up with me full time :)

Melody  Feb 18, 2019
 #6
avatar+6045 
+1

btw there's a small error in your answer.. I'll let you find it!

Rom  Feb 18, 2019
 #7
avatar+105468 
-1

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   laugh

Melody  Feb 18, 2019
 #8
avatar+23273 
+3

If

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

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

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

 

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

 

laugh

 Feb 18, 2019

25 Online Users

avatar