Question 1110100
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The domain of f(x) is all values except 4/5.<br>
That means the range of f^-1(x) is all values except 4/5.<br>
To find f^-1(x), switch x and y in the given function and solve for the new y.<br>
{{{x = y/(5y-4)}}}
{{{x(5y-4) = y}}}
{{{5xy-4x = y}}}
{{{5xy-y = 4x}}}
{{{(5x-1)y = 4x}}}
{{{y = 4x/(5x-1)}}}<br>
{{{f^-1(x) = (4x)/(5x-1)}}}<br>
The domain of f^-1(x) is all values except 1/5.<br>
Graphing the function and the inverse confirm these results.<br>
The graph of f(x) (red) has a vertical asymptote at x=4/5 and a horizontal asymptote at y=1/5 (green).<br>
{{{graph(400,400,-2,2,-2,2,x/(5x-4),1/5)}}}<br>
The graph of f^-1(x) (red) has a vertical asymptote at x=1/5 and a horizontal asymptote at y = 4/5 (green).<br>
{{{graph(400,400,-2,2,-2,2,(4x)/(5x-1),4/5)}}}<br>
ANSWER:
f^-1(x) = 4x/(5x-1)
Domain: x not = 1/5
Range: y not = 4/5