What is the inverse of f(x)=(x+8)^2,x>=-8 ?

The inverse of f(x)=(x+8)^2,x>=-8 is sqrt(x)-8,-sqrt(x)-8

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inverse of f(x)=(x+8)^2,x>=-8

inverse

f (x) = (x + 8)^{2}, x \geq - 8

x - 8, - x - 8

Solution steps

y = (x + 8)^{2}

Replace

x

with

y

x = (y + 8)^{2}

Solve for

y, x = (y + 8)^{2}

x - 8, - x - 8

f (x) = \frac{( 6 - 6 x )}{( 3 - 7 x )}

y = 2 (x - 5)^{2} - 6

f (x) = \frac{2 ( x + 1 )}{4 - x}

g (x) = \frac{5 x - 1}{4 x + 7}

What is the inverse of f(x)=(x+8)^2,x>=-8 ?
The inverse of f(x)=(x+8)^2,x>=-8 is sqrt(x)-8,-sqrt(x)-8