What is the inverse of f(x)=sqrt(x)+sqrt(x-1) ?

The inverse of f(x)=sqrt(x)+sqrt(x-1) is (x^4+2x^2+1)/(4x^2)

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inverse of f(x)=sqrt(x)+sqrt(x-1)

inverse

f (x) = x + x - 1

\frac{x ^{4} + 2 x ^{2} + 1}{4 x ^{2}}

Solution steps

y = x + x - 1

Replace

x

with

y

x = y + y - 1

Solve for

y, x = y + y - 1

\frac{x ^{4} + 2 x ^{2} + 1}{4 x ^{2}}

f (x) = \frac{3}{2} cos (5 x - \frac{π}{4})

f (x) = \frac{- 3 x - 4}{x - 2}

f (x) = 5 - 2 x - 3, - 2 \leq x < 2

f (x) = 2 - 7 x

f (x) = 3 x - 8

What is the inverse of f(x)=sqrt(x)+sqrt(x-1) ?
The inverse of f(x)=sqrt(x)+sqrt(x-1) is (x^4+2x^2+1)/(4x^2)