AMC 10 Daily Practice - Polynomial Functions

Complete problem set with solutions and individual problem pages

Problem 2 Easy

Let the function f(x) = x^3 + 3x^2 + 6x + 14 and suppose that f(a) = 1, f(b) = 19. What is the value of a + b?

  • A.

    2

  • B.

    1

  • C.

    0

  • D.

    -1

  • E.

    -2

Answer:E

Given f(x) = x^3 + 3x^2 + 6x + 14 = (x + 1)^3 + 3(x + 1) + 10, let g(y) = y^3 + 3y, then g(y) is an odd function and is strictly increasing.

Since f(a) = (a + 1)^3 + 3(a + 1) + 10 = 1, \\f(b) = (b + 1)^3 + 3(b + 1) + 10 = 19, we have g(a + 1) = -9,\quad g(b + 1) = 9,\quad g(-b - 1) = -9.

Therefore, g(a + 1) = g(-b - 1),

which implies a + 1 = -b - 1 \Rightarrow a + b = -2.

Answer: E.

Related Problems
Problem 1 AMC 10 Daily Practice - Polynomial Functions
Problem 3 AMC 10 Daily Practice - Polynomial Functions
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