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

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

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

inverse

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

2 + x + 4, 2 - x + 4

Solution steps

y = x^{2} - 4 x

Replace

x

with

y

x = y^{2} - 4 y

Solve for

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

2 + x + 4, 2 - x + 4

x - y = - 1

(- 2, 7), (- 2, - 2)

x^{2} + 8 x

f (x) = ln (x) + 7

5 x - 2 y = 4, (2, - 4)

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