f (x) = (x − a)3 + (x − b)3 + (x − c)3 , a < b < c.
The number of real roots of f (x) = 0 is
(A) 3; (B) 2; (C) 1; (D) 0.
Please guide me with this question..
$\displaystyle \lim_{x\to -\infty}=-\infty, \ \lim_{x\to\infty}=\infty\Rightarrow$ f has at least one real root.
$\displaystyle f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2>0, \forall x\in\mathbf{R}\Rightarrow$ f is strictly increasing, so f is injective. Therefore, f has only one real root.