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

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

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

inverse

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

x + 8, - x + 8

Solution steps

y = x^{2} - 8

Replace

x

with

y

x = y^{2} - 8

Solve for

y, x = y^{2} - 8

x + 8, - x + 8

0.1

f (x) = \frac{5 x + 2}{3 x - 1}

+ x^{3} - 5

3 x^{2} - 12 x + 10

x - \frac{5}{2}

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