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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = 0.5 x + 6.4

f (x) = 5 (x - 3)

[1]

- x + 2

f (x) = lo g_{2} (x - 12)

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