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

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

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

inverse

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

\frac{1}{x - 2}, - \frac{1}{x - 2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1}{x - 2}, - \frac{1}{x - 2}

f (x) = 6 x + 15

\frac{4}{s}

10 - 3

x^{(2)} - 5 x - 7

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

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