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

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

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inverse of f(x)= 1/7 x^2-1

inverse

f (x) = \frac{1}{7} x^{2} - 1

7 x + 7, - 7 x + 7

Solution steps

y = \frac{1}{7} x^{2} - 1

Replace

x

with

y

x = \frac{1}{7} y^{2} - 1

Solve for

y, x = \frac{1}{7} y^{2} - 1

7 x + 7, - 7 x + 7

f (x) = \frac{8}{( x + 3 )} - 2

f (x) = \frac{( x + 12 )}{x - 8}

y = 2 - x^{2}

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

f (x) = 3 x^{4} + 4 x^{3} + 6 x^{2} - 4

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