What is the inverse of y=(60x)/(1x^2+100) ?

The inverse of y=(60x)/(1x^2+100) is (10(sqrt(-x^2+9)+3))/x ,(10(-sqrt(-x^2+9)+3))/x

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inverse of y=(60x)/(1x^2+100)

inverse

y = \frac{60 x}{1 x ^{2} + 100}

\frac{10 ( - x ^{2} + 9 + 3 )}{x}, \frac{10 ( - - x ^{2} + 9 + 3 )}{x}

Solution steps

y = \frac{60 x}{1 \cdot x ^{2} + 100}

Replace

x

with

y

x = \frac{60 y}{1 \cdot y ^{2} + 100}

Solve for

y, x = \frac{60 y}{1 y ^{2} + 100}

\frac{10 ( - x ^{2} + 9 + 3 )}{x}, \frac{10 ( - - x ^{2} + 9 + 3 )}{x}

21 x^{2} + 5 x - 84

f (x) = \frac{x}{2} + \frac{x - 2 ( x + 2 )}{4}

f (x) = (x + 1)^{0.5}

f (x) = f (x) = - 4 x - 1

f (x) = - 2 x + 11

What is the inverse of y=(60x)/(1x^2+100) ?
The inverse of y=(60x)/(1x^2+100) is (10(sqrt(-x^2+9)+3))/x ,(10(-sqrt(-x^2+9)+3))/x