What is the inverse of (x-7)/(x^2) ?

The inverse of (x-7)/(x^2) is (1+sqrt(1-28x))/(2x),(1-sqrt(1-28x))/(2x)

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inverse of (x-7)/(x^2)

inverse

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

\frac{1 + 1 - 28 x}{2 x}, \frac{1 - 1 - 28 x}{2 x}

Solution steps

y = \frac{x - 7}{x ^{2}}

Replace

x

with

y

x = \frac{y - 7}{y ^{2}}

Solve for

y, x = \frac{y - 7}{y ^{2}}

\frac{1 + 1 - 28 x}{2 x}, \frac{1 - 1 - 28 x}{2 x}

y = \frac{9 x - 3}{9 x + 7}

- \frac{1}{2} x + \frac{3}{4}

f (x) = 3 + ln (- 3 + 2 x)

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

What is the inverse of (x-7)/(x^2) ?
The inverse of (x-7)/(x^2) is (1+sqrt(1-28x))/(2x),(1-sqrt(1-28x))/(2x)