What is the inverse of 1-3/(x^2-1) ?

The inverse of 1-3/(x^2-1) is sqrt((x-4)/(x-1)),-sqrt((x-4)/(x-1))

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inverse of 1-3/(x^2-1)

inverse

1 - \frac{3}{x ^{2} - 1}

\frac{x - 4}{x - 1}, - \frac{x - 4}{x - 1}

Solution steps

y = 1 - \frac{3}{x ^{2} - 1}

Replace

x

with

y

x = 1 - \frac{3}{y ^{2} - 1}

Solve for

y, x = 1 - \frac{3}{y ^{2} - 1}

\frac{x - 4}{x - 1}, - \frac{x - 4}{x - 1}

- 3 x + 2 - 6

f (x) = 2 x^{2} - 8 x + 10

f (x) = 9 (\frac{x}{2}) - 8

f (x) = lo g_{2} (x) + 3

\frac{5 - x}{3 x - 4}

What is the inverse of 1-3/(x^2-1) ?
The inverse of 1-3/(x^2-1) is sqrt((x-4)/(x-1)),-sqrt((x-4)/(x-1))