What is the inverse of f(x)=sqrt(x^2-8x),x<= 0 ?

The inverse of f(x)=sqrt(x^2-8x),x<= 0 is 4+sqrt(x^2+16),4-sqrt(x^2+16)

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inverse of f(x)=sqrt(x^2-8x),x<= 0

inverse

f (x) = x^{2} - 8 x, x \leq 0

4 + x^{2} + 16, 4 - x^{2} + 16

Solution steps

y = x^{2} - 8 x

Replace

x

with

y

x = y^{2} - 8 y

Solve for

y, x = y^{2} - 8 y

4 + x^{2} + 16, 4 - x^{2} + 16

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

f (x) = \frac{4 x - 12}{( x - 2 ) ^{2}}

\frac{- 5 x + 1}{5 + 6 x}

f (x) = 1 - x

x - 2 + 5

What is the inverse of f(x)=sqrt(x^2-8x),x<= 0 ?
The inverse of f(x)=sqrt(x^2-8x),x<= 0 is 4+sqrt(x^2+16),4-sqrt(x^2+16)