What is the inverse of y= 6/(x^2+1) ?

The inverse of y= 6/(x^2+1) is sqrt((6-x)/x),-sqrt((6-x)/x)

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inverse of y= 6/(x^2+1)

inverse

y = \frac{6}{x ^{2} + 1}

\frac{6 - x}{x}, - \frac{6 - x}{x}

Solution steps

y = \frac{6}{x ^{2} + 1}

Replace

x

with

y

x = \frac{6}{y ^{2} + 1}

Solve for

y, x = \frac{6}{y ^{2} + 1}

\frac{6 - x}{x}, - \frac{6 - x}{x}

\frac{( ln ( x - 1 ) )}{x - 1}

(- 4, - 3), (6, 5)

f (x) = - 3 x + 12

f (x) = lo g_{x} (x^{2})

y - 2 = - (x - 4)

What is the inverse of y= 6/(x^2+1) ?
The inverse of y= 6/(x^2+1) is sqrt((6-x)/x),-sqrt((6-x)/x)