What is the inverse of f(x)= 1/(sqrt(x^2+7)) ?

The inverse of f(x)= 1/(sqrt(x^2+7)) is (sqrt(1-7x^2))/x ,-(sqrt(1-7x^2))/x

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inverse of f(x)= 1/(sqrt(x^2+7))

inverse

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

\frac{1 - 7 x ^{2}}{x}, - \frac{1 - 7 x ^{2}}{x}

Solution steps

y = \frac{1}{x ^{2} + 7}

Replace

x

with

y

x = \frac{1}{y ^{2} + 7}

Solve for

y, x = \frac{1}{y ^{2} + 7}

\frac{1 - 7 x ^{2}}{x}, - \frac{1 - 7 x ^{2}}{x}

f (x) = 0.5 x^{2} - 8 x + 35

f (x) = - x^{4} - 7 x^{3} + 3 x + 8

5 - x + 1

\frac{x ^{3}}{( x - 1 ) ^{2}}

r (x) = \frac{5 x ^{3}}{x ^{3} + 2 x ^{2} + 5 x}

What is the inverse of f(x)= 1/(sqrt(x^2+7)) ?
The inverse of f(x)= 1/(sqrt(x^2+7)) is (sqrt(1-7x^2))/x ,-(sqrt(1-7x^2))/x