What is the inverse of y=(x^2+1)/(x^2+3) ?

The inverse of y=(x^2+1)/(x^2+3) is sqrt((1-3x)/(x-1)),-sqrt((1-3x)/(x-1))

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inverse of y=(x^2+1)/(x^2+3)

inverse

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

\frac{1 - 3 x}{x - 1}, - \frac{1 - 3 x}{x - 1}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 - 3 x}{x - 1}, - \frac{1 - 3 x}{x - 1}

f (x) = (x - 7)^{3} + 3

S (x) = 6 x^{2}

f (x) = \frac{( 7 + 4 x )}{25 - 8 x}

f (x) = 2 - lo g_{10} (x)

f (x) = 3 lo g_{4} (x + 2) - 3

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