What is the inverse of sqrt(10-2x+3\sqrt{x-1)} ?

The inverse of sqrt(10-2x+3\sqrt{x-1)} is (-4x^2+49+3sqrt(-8x^2+73))/8 ,(-4x^2+49-3sqrt(-8x^2+73))/8

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inverse of sqrt(10-2x+3\sqrt{x-1)}

inverse

10 - 2 x + 3 x - 1

\frac{- 4 x ^{2} + 49 + 3 - 8 x ^{2} + 73}{8}, \frac{- 4 x ^{2} + 49 - 3 - 8 x ^{2} + 73}{8}

Solution steps

y = 10 - 2 x + 3 x - 1

Replace

x

with

y

x = 10 - 2 y + 3 y - 1

Solve for

y, x = 10 - 2 y + 3 y - 1

\frac{- 4 x ^{2} + 49 + 3 - 8 x ^{2} + 73}{8}, \frac{- 4 x ^{2} + 49 - 3 - 8 x ^{2} + 73}{8}

\frac{x}{3}

f (x) = \frac{( 1 - 3 x )}{( 2 x + 1 )}

lo g_{e} ((cos (x))^{2})

f (x) = x^{3} - 1, x > \frac{π}{2}

y = x + 1

What is the inverse of sqrt(10-2x+3\sqrt{x-1)} ?
The inverse of sqrt(10-2x+3\sqrt{x-1)} is (-4x^2+49+3sqrt(-8x^2+73))/8 ,(-4x^2+49-3sqrt(-8x^2+73))/8