Question

D(f) -?
$f(x) = {x}^{2} + 2x \\ f(b - 1)$
​

Answer 1

$f(x) = (b - 1) {}^{2} + 2(b - 1) \\ f(x) = b {}^{2} - 2b + 1 + 2b - 2 \\ f(x) = b {}^{2} - 1$

D(f)

y принадлежит [-1;+бесконечность)