What is the answer to solvefor x,f^{-1}(x)=sqrt(2x) ?

The answer to solvefor x,f^{-1}(x)=sqrt(2x) is x=2f^2 {f>0},x=0 {f 0}

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solvefor x,f^{-1}(x)=sqrt(2x)

solve for

x, f^{- 1} (x) = 2 x

x = 2 f^{2} {f > 0}, x = 0 {f < 0 or f > 0}

Solution steps

f^{- 1} (x) = 2 x

Expand

f^{- 1} x : \frac{x}{f}

\frac{x}{f} = 2 x

Square both sides:

\frac{x ^{2}}{f ^{2}} = 2 x

\frac{x ^{2}}{f ^{2}} = 2 x

Solve

\frac{x ^{2}}{f ^{2}} = 2 x : x = 2 f^{2}, x = 0

x = 2 f^{2}, x = 0

Verify Solutions:

x = 2 f^{2} {f > 0}, x = 0 {f < 0 or f > 0}

The solutions are

x = 2 f^{2} {f > 0}, x = 0 {f < 0 or f > 0}

x + x - 12 = 2

5 = x^{2} + 3^{2}

x^{2} - 16 x = 12

x^{2} + 2 x - 1 = 5

x + 5 - x - 2 = 1

What is the answer to solvefor x,f^{-1}(x)=sqrt(2x) ?
The answer to solvefor x,f^{-1}(x)=sqrt(2x) is x=2f^2 {f>0},x=0 {f<0\lor f>0}