f(x)= 4x−1f(x) = \; \frac{4}{x} - 1f(x)=x4−1. Let y = f(x)
y= 4−xx ⟹ xy+x=4y = \; \frac{4 - x}{x} \implies xy + x = 4y=x4−x⟹xy+x=4
x(y+1)=4∴x= 4y+1x(y + 1) = 4 \therefore x = \; \frac{4}{y + 1}x(y+1)=4∴x=y+14
f−1(7)= 47+1= 12f^{-1}(7) = \; \frac{4}{7 + 1} = \; \frac{1}{2}f−1(7)=7+14=21
