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 $f(x) = -3x^3 + 3x^2 + 6x\,$ .

Last modified on 27 October 2009, at 03:39