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

9

$f$ is a cubic, so must be of the form $f(x)=a{x}^3+b{x}^2+cx+d$.

Therefore, $f(0)=d=0$, so now $f(x)=a{x}^3+b{x}^2+cx$

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