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

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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