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

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

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

inverse

f (x) = \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) = \frac{2}{π} \cdot sin (r)

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

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

f (x) = \frac{4}{5} x - 6

f (x) = arcsin (6 x^{5})

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