Let $f(x) = x^2 + ax + b$ and $g(x) = f(x + 1) - f(x - 1)$. If $f(x) \ge 0$ for all real $x$ and $g(15) = 48$, then the smallest possible value of $b$ is:
Answer and solution
Correct answer: C
Solution
$f(x+1) = x^2 + 2x + 1 + ax + a + b$
$f(x-1) = x^2 - 2x + 1 + ax - a + b$
$g(x) = f(x+1) - f(x-1) = 4x + 2a$
$g(15) = 60 + 2a = 48 \Rightarrow a = -6$
$f(x) = x^2 - 6x + b = (x - 3)^2 + b - 9$
Its least value is $b - 9$ (at $x = 3$). For $f(x) \ge 0$ always: $b - 9 \ge 0 \Rightarrow b \ge 9$
Smallest $b = 9$
$f(x-1) = x^2 - 2x + 1 + ax - a + b$
$g(x) = f(x+1) - f(x-1) = 4x + 2a$
$g(15) = 60 + 2a = 48 \Rightarrow a = -6$
$f(x) = x^2 - 6x + b = (x - 3)^2 + b - 9$
Its least value is $b - 9$ (at $x = 3$). For $f(x) \ge 0$ always: $b - 9 \ge 0 \Rightarrow b \ge 9$
Smallest $b = 9$