What is the inverse of f(x)=x^2-16x,x>= 8 ?

The inverse of f(x)=x^2-16x,x>= 8 is 8+sqrt(x+64),8-sqrt(x+64)

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inverse of f(x)=x^2-16x,x>= 8

inverse

f (x) = x^{2} - 16 x, x \geq 8

8 + x + 64, 8 - x + 64

Solution steps

y = x^{2} - 16 x

Replace

x

with

y

x = y^{2} - 16 y

Solve for

y, x = y^{2} - 16 y

8 + x + 64, 8 - x + 64

f (x) = \frac{x - 7}{( x - 7 ) ( x - 5 )}

f (x) = 3 (x - 1)

f (x) = \frac{1}{2} x^{2} + 6 x + 7

2 x + 3

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

What is the inverse of f(x)=x^2-16x,x>= 8 ?
The inverse of f(x)=x^2-16x,x>= 8 is 8+sqrt(x+64),8-sqrt(x+64)