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**+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 / a^{m}

Therefore..... 3^{-2} = 1 / 3^{2} = 1 / 9

So......[ f^{-1}(2 ) ] ^{-2} = 1 / 9

And that's it..!!!!

CPhill Jul 19, 2017