What is the inverse of 1/((x^2-9)^{1/2)} ?

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

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inverse of 1/((x^2-9)^{1/2)}

inverse

\frac{1}{( x ^{2} - 9 ) ^{\frac{1}{2}}}

\frac{( 1 + 9 x ^{2} ) ^{\frac{1}{2}}}{x}, - \frac{( 1 + 9 x ^{2} ) ^{\frac{1}{2}}}{x}

Solution steps

y = \frac{1}{( x ^{2} - 9 ) ^{\frac{1}{2}}}

Replace

x

with

y

x = \frac{1}{( y ^{2} - 9 ) ^{\frac{1}{2}}}

Solve for

y, x = \frac{1}{( y ^{2} - 9 ) ^{\frac{1}{2}}}

\frac{( 1 + 9 x ^{2} ) ^{\frac{1}{2}}}{x}, - \frac{( 1 + 9 x ^{2} ) ^{\frac{1}{2}}}{x}

f (x) = 2^{x + 2}

f (x) = 740.85 x - 450.2

f (x) = lo g_{e} (5 x + 1) - 7

f (x) = \frac{( x + 2 )}{x - 2}

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

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