What is the inverse of 1/(sqrt(x^2+9)) ?

The inverse of 1/(sqrt(x^2+9)) is (sqrt(1-9x^2))/x ,-(sqrt(1-9x^2))/x

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inverse of 1/(sqrt(x^2+9))

inverse

\frac{1}{x ^{2} + 9}

\frac{1 - 9 x ^{2}}{x}, - \frac{1 - 9 x ^{2}}{x}

Solution steps

y = \frac{1}{x ^{2} + 9}

Replace

x

with

y

x = \frac{1}{y ^{2} + 9}

Solve for

y, x = \frac{1}{y ^{2} + 9}

\frac{1 - 9 x ^{2}}{x}, - \frac{1 - 9 x ^{2}}{x}

lo g_{10} (- 8.5)

v (x) = x^{3} + 7

- 4.5

\frac{( 3 x + 2 )}{( 2 x + 5 )}

What is the inverse of 1/(sqrt(x^2+9)) ?
The inverse of 1/(sqrt(x^2+9)) is (sqrt(1-9x^2))/x ,-(sqrt(1-9x^2))/x