Solutions To Mathematics Textbooks/Calculus Early Transcendentals (6th) (0495011665)/Chapter 1.2

9
$$f\,$$ is a cubic, so must be of the form $$f(x) = ax^3 + bx^2 + cx + d\,$$.

Therefore, $$f(0) = d = 0\,$$, so now $$f(x) = ax^3 + bx^2 + cx\,$$

Looking at other values:

$$f(1) = a + b + c = 6\,$$

$$f(-1) = -a + b -c = 0\,$$

$$f(2) = 8a + 4b + 2c = 0\,$$

Try to solve simulatenously:

$$(a + b + c) + (-a + b -c) = 6\,$$

$$2b = 6\,$$, thus $$b = 3\,$$.

$$(a + b + c) - (-a + b -c) = 6\,$$

$$2a + 2c = 6\,$$, thus $$a + c = 3\,$$.

Thus,

$$a + c = 3\,$$

$$8a + 2c = -12\,$$

Solve simultaneous equation:

$$(2a + 2c) - (8a + 2c) = 18\,$$

$$-6a = 18\,$$ therefore $$a = -3\,$$.

If $$b = 3, a = -3\,$$ then:

$$f(1) = -3 + 3 + c = 6\,$$, thus $$c = 6\,$$.

Therefore, we have $$a = -3, b=3, c=6\,$$ and our function is given by $$f(x) = -3x^3 + 3x^2 + 6x\,$$.