$f(x) = 20x - 4 \\ substitute \: y \: for \: f(x) \\ y = 20x - 4 \\ interchange \: x \: and \: y \\ x = 20y - 4 \\ swap \: the \: sides \: of \: the \: equation \\ 20y - 4 = x \\ move \: the \: constant \: to \: the \: right \: hand \\ 20y = x + 4 \\ divide \: both \: sides \: by \: 20 \\ y = \frac{1}{20} x + \frac{1}{5} \\ substitute \: f {}^{ - 1} (x) \: for \: y \\ f {}^{ - 1} (x) = \frac{1}{20} x + \frac{1}{5}$

The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.

Thus, domain of f(x): x∈R = range of f¯¹(x)

and range of f(x): x∈R =domain of f¯¹(x)

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2 years ago

*MARKING BRAINLIEST*

wlad13 [49]

Answer:

$(x+y+z)^2$