What is the inverse of f(x)=((x^2-5))/(7x^2) ?

The inverse of f(x)=((x^2-5))/(7x^2) is sqrt(-5/(7x-1)),-sqrt(-5/(7x-1))

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inverse of f(x)=((x^2-5))/(7x^2)

inverse

f (x) = \frac{( x ^{2} - 5 )}{7 x ^{2}}

- \frac{5}{7 x - 1}, - - \frac{5}{7 x - 1}

Solution steps

y = \frac{x ^{2} - 5}{7 x ^{2}}

Replace

x

with

y

x = \frac{y ^{2} - 5}{7 y ^{2}}

Solve for

y, x = \frac{y ^{2} - 5}{7 y ^{2}}

- \frac{5}{7 x - 1}, - - \frac{5}{7 x - 1}

f (x) = 2 \cdot x^{2}

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

f (x) = \frac{3 x}{( x ^{2} - 4 )}

f (x) = 2 - 6 x^{2}, x < 0

\frac{5 x + 4}{4 x - 2}

What is the inverse of f(x)=((x^2-5))/(7x^2) ?
The inverse of f(x)=((x^2-5))/(7x^2) is sqrt(-5/(7x-1)),-sqrt(-5/(7x-1))