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

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

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

inverse

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

\frac{1}{b ^{2} - 1}, - \frac{1}{b ^{2} - 1}

Solution steps

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

Replace

b

with

y

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

Solve for

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

\frac{1}{b ^{2} - 1}, - \frac{1}{b ^{2} - 1}

f (x) = \frac{3 x + 8}{x - 7}

f (x) = 0.12 x + 50

f (x) = ln (x) ln (x)

- x^{2} + 6 x + 4

\frac{x - 7}{x + 2}

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