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If \(f(x)=\frac{16}{5+3x}\) , what is the value of \(\left[f^{-1}(2)\right]^{-2}\) ?

Guest Jul 19, 2017
edited by Guest  Jul 19, 2017
 #1
avatar+90088 
+1

 

Let's first write this as

 

y = 16 / [ 5 + x ]

 

We need to find the inverse ....the idea is to isolate x and then "exchange" x and y

 

Multiply both sides by [ 5 + x ]

 

[ 5 + x ] y =  16     now....divide both sides by y

 

5 + x  = 16/y        subtract  5 from both sides

 

x =  16/y  - 5    "  exchange " x and y

 

y = 16/x - 5       get a common denominator on the right

 

y =  16/x  -  5x / x

 

y =  [ 16 - 5x ] / x      and for y, write f-1(x)

 

f-1(x) =  [ 16 - 5x ] / x

 

Now....we want to first find  [ f-1(2 ) ]

 

This means that we want to put 2 into the inverse and evaluate it.....so we have

 

[ 16 -  5(2) ] / 2  =   6/2  = 3

 

So..... [ f-1(2 ) ] =  3

 

Now we want to find   [ f-1(2 ) ] -2    which is just [ 3 ]-2

 

Remember that a-m  =  1 / am

 

Therefore..... 3-2  =  1 / 32   =   1 / 9

 

So......[ f-1(2 ) ] -2 =   1 / 9

 

And that's it..!!!!

 

 

 

cool cool cool

CPhill  Jul 19, 2017
edited by CPhill  Jul 19, 2017
 #2
avatar+738 
+6

Thanks so much!!!! You're the best

MIRB16  Jul 19, 2017
 #3
avatar+90088 
0

 

Guest...if you're still on....check my answer again...I made a slight mistake the first time....!!!

 

 

cool cool cool

CPhill  Jul 19, 2017

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