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

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

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

inverse

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

x + 4, - x + 4

Solution steps

y = (x - 4)^{2}

Replace

x

with

y

x = (y - 4)^{2}

Solve for

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

x + 4, - x + 4

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

f (x) = (x - 2)^{2} - 7

f (x) = \frac{2}{x - 3} + 5

f (x) = x^{3} - 3 x + 8

f (x) = cos^{2} (x)

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