Question

Find $p\left(3\right)\text{and}p\left(-1\right)$ for the function $p\left(x\right)=-x^3+x^2-x$.

Short answer

The value of $p\left(3\right)=-21\text{and}p\left(-1\right)=3$.

Step-by-step solution

Step 1. Write down the given information.

The given function is $p\left(x\right)=-x^3+x^2-x$.

Step 2. Calculation.

Evaluating $p\left(3\right)\text{and}p\left(-1\right)$ by plugging $x=3\text{and}x=-1$ in given function $p\left(x\right)=-x^3+x^2-x$.

$\begin{array}{c}p\left(x\right)=-x^3+x^2-x\mathrm{....}\left(\text{Given}\right)\\ p\left(3\right)=-{\left(3\right)}^3+{\left(3\right)}^2-\left(3\right)\\ =-21\\ \text{Now,}p\left(-1\right)=-{\left(-1\right)}^3+{\left(-1\right)}^2-\left(-1\right)\\ =3\end{array}$

Step 3. Conclusion.

The value of $p\left(3\right)=-21\text{and}p\left(-1\right)=3$.