Method 1:

`f(x) = kx -2`

`{\rm let } f(x)^{-1}= y`

`f(y) = x`

`ky - 2 = x`

`ky = x + 2`

`y =\frac{{x + 2}}{k}`

`f^{-1}(x) = \frac{{x + 2}}{k}`

Compare the answer

`f^{ -1} (x) = \frac{{x + 2}}{k}{\rm with }f^{ - 1} (x) = \frac{{x + 2}}{5}`

`k = 5`

Method 2:

f

^{-1}(x)=(x+2)/5

let f(x) = y

f

^{-1}(y) = x

(y + 2)/5 = x

(y + 2) = 5x

y = 5x -2

Substitute y with f(x) (because f(x) = y)

f(x) = 5x -2

Compare with f(x) = kx -2 (that given in the question), we get

k = 5