What is the inverse of f(x)=(x-4)^2-1,x>= 4 ?

The inverse of f(x)=(x-4)^2-1,x>= 4 is sqrt(x+1)+4,-sqrt(x+1)+4

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inverse of f(x)=(x-4)^2-1,x>= 4

inverse

f (x) = (x - 4)^{2} - 1, x \geq 4

x + 1 + 4, - x + 1 + 4

Solution steps

y = (x - 4)^{2} - 1

Replace

x

with

y

x = (y - 4)^{2} - 1

Solve for

y, x = (y - 4)^{2} - 1

x + 1 + 4, - x + 1 + 4

f (x) = sin (16 - x^{2})

sin (\frac{2}{6})

f (x) = x^{2} - 4 x - 16

lo g_{6} (4 x + 4)

f (x) = \frac{- 4 x + 1}{9 + 5 x}

What is the inverse of f(x)=(x-4)^2-1,x>= 4 ?
The inverse of f(x)=(x-4)^2-1,x>= 4 is sqrt(x+1)+4,-sqrt(x+1)+4