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

## 9

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

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

Looking at other values:

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

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

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

Try to solve simulatenously:

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

${\displaystyle 2b=6\,}$, thus ${\displaystyle b=3\,}$.

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

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

Thus,

${\displaystyle a+c=3\,}$

${\displaystyle 8a+2c=-12\,}$

Solve simultaneous equation:

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

${\displaystyle -6a=18\,}$ therefore ${\displaystyle a=-3\,}$.

If ${\displaystyle b=3,a=-3\,}$ then:

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

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