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

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

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

inverse

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

2 x + 1, - 2 x + 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

2 x + 1, - 2 x + 1

f (x) = \frac{1}{lo g _{10} ( x + 1 )}

\frac{3 x + 1}{- 5 x - 2}

f (x) = 6 + x - 8

f (x) = \frac{( 5 x - 2 )}{9}

f (x) = \frac{x + 2}{x + 12}

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