Question

If g(x) = x^2 + 6x with x ≥ -3, find g ^-1(0).

Answer 1

Answer: 0

Step-by-step explanation:

g(x) = x² + 6x   ; x ≥ -3

To find the inverse, swap the x's and y's and solve for "y":

    x = y² + 6y

x + 9 = y² + 6y + 9   add 9 to both sides to create a perfect square

x + 9 = (y + 3)²

$+/-\sqrt{x+9}$ = y + 3   take square root of both sides

$-3 +/-\sqrt{x+9}$ = y    ; y ≥ -3

g⁻¹(0) = $-3 +/-\sqrt{0+9}$

        = $-3 +/-\sqrt{9}$

        = -3 ± 3

        = -3 + 3   ,   -3 - 3

        =     0      ,      -6

since the restriction is: y ≥ -3, then -6 is not valid