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avatar+22 

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

 

Thanks!!!

 

laughlaughlaughlaughlaughlaughlaugh

 Mar 11, 2019
 #1
avatar+111394 
0

f(0)  =   a/2

 

To find the inverse, we have that

 

y =   a / [x + 2]

y  [ x + 2 ] = a

yx + 2y  =  a

yx =  a - 2y

x  =  [ a - 2y] / y       swap  x and y

y = [ a - 2x ] /x     =  the inverse = f-1(x)

 

So

 

f-1 (3a)  =  [ a - 2(3a) ] / [3a ]      =   [ -5a ] / 3a  =  -5/3

 

So  if f(0)  = f-1(3a).....then

 

a/ 2   =  -5/3

 

a = -10/3

 

 

cool cool cool

 Mar 11, 2019
 #2
avatar+25255 
+1

Inverse Functions

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

solve for the value of  \(a\) so that

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

 

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

 

laugh

 Mar 11, 2019

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