The function f(x)=(x+4)^2 + 3 is not one-to-one. Identify a restricted domain that makes the function one-to-one, ad find the inverse function.
+129838 f(x)=(x+4)^2 + 3
The vertex of this function is (-4, 3).........it will be one-to-one if we resrict the domain to either (-inf, -4) or (-4, inf).....x = -4 could belong to either original domain and to either domain of the inverse
To find the inverse write "y" for f(x)
y = (x + 4)^2 + 3 subtract 3 from both sides
y - 3 = ( x + 4)^2 take the pos/neg square roots of both sides
±√ [y - 3] = x + 4 subtract 4 from both sides
±√ [y - 3] - 4 = x "swap" x and y
±√ [x - 3] - 4 = y for y, write f-1(x)
±√ [x - 3] - 4 = f-1 (x)
When we restrict the original domain to (-inf, -4), -√ [x - 3] - 4 is the inverse
When we restrict the original domain to (-4, inf ) , √ [x - 3] - 4 is the inverse
