Question

Can anyone help me please

Answer 1

To find $f^{-1} (x)$, you can switch the x and the y(which is f(x)).

So instead of $f(x)=\frac{-7}{4x-5}$, it will be:

$x = \frac{-7}{4y-5}$

Now you need to find "y".

You first multiply (4y - 5) on both sides

$x(4y - 5) = -7$

Next divide "x" on both sides

$4y - 5 = \frac{-7}{x}$

Add 5 on both sides

$4y = \frac{-7}{x} +5$

Divide 4 on both sides to get "y" by itself

$y = \frac{7}{4x} +\frac{5}{4}$

$f^{-1} = \frac{7}{4x} +\frac{5 }{4}$