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..

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- March 3rd 2009, 08:37 AMchamprockCube root
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.. - March 3rd 2009, 09:53 AMred_dog
f has at least one real root.

f is strictly increasing, so f is injective. Therefore, f has only one real root.