Question

If $p\left(x\right)=2+\frac{x}{2}+x^2-\frac{x^3}{3}$, then $p(-1)$ is :

• A. $\frac{15}{6}$
• B. $\frac{17}{6}$
• C. $\frac{1}{6}$
• D. $\frac{13}{6}$

Answer 1

The correct option is B $\frac{17}{6}$

$p\left(x\right)=2+\frac{x}{2}+x^2-\frac{x^3}{3}$

Putting $x=-1$

$p(-1)=2+\frac{(-1)}{2}+{(-1)}^2-\frac{{(-1)}^3}{3}=2-\frac{1}{2}+1+\frac{1}{3}=\frac{5}{2}+\frac{1}{3}=\frac{17}{6}$