What is the inverse of (2x^2+5)/(x^2-1) ?

The inverse of (2x^2+5)/(x^2-1) is sqrt((5+x)/(x-2)),-sqrt((5+x)/(x-2))

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inverse of (2x^2+5)/(x^2-1)

inverse

\frac{2 x ^{2} + 5}{x ^{2} - 1}

\frac{5 + x}{x - 2}, - \frac{5 + x}{x - 2}

Solution steps

y = \frac{2 x ^{2} + 5}{x ^{2} - 1}

Replace

x

with

y

x = \frac{2 y ^{2} + 5}{y ^{2} - 1}

Solve for

y, x = \frac{2 y ^{2} + 5}{y ^{2} - 1}

\frac{5 + x}{x - 2}, - \frac{5 + x}{x - 2}

f (x) = 2 x^{2} - 12 x + 13

h (x) = \frac{1 + 2 x}{7 + x}

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

f (x) = 100

f (x) = - 5 x^{2} - 9, x \geq 0

What is the inverse of (2x^2+5)/(x^2-1) ?
The inverse of (2x^2+5)/(x^2-1) is sqrt((5+x)/(x-2)),-sqrt((5+x)/(x-2))