What is the inverse of f(x)=x^2-16,x<= 0 ?

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

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inverse of f(x)=x^2-16,x<= 0

inverse

f (x) = x^{2} - 16, x \leq 0

x + 16, - x + 16

Solution steps

y = x^{2} - 16

Replace

x

with

y

x = y^{2} - 16

Solve for

y, x = y^{2} - 16

x + 16, - x + 16

2 x - 3 y = 6

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

7 y - 8 = - 2 (2 - x)

- \frac{3}{2}

(- 1, - 10), (- 9, - 4)

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