Question

Find $p\left(4\right)\text{and}p\left(-2\right)$ for the function $p\left(x\right)=\frac{1}{8}{x}^3-\frac{1}{4}{x}^2-\frac{1}{2}x+5$.

Short answer

The value of $p\left(4\right)=7\text{and}p\left(-2\right)=4$.

Step-by-step solution

Step 1. Write down the given information.

The given function is $p\left(x\right)=\frac{1}{8}{x}^3-\frac{1}{4}{x}^2-\frac{1}{2}x+5$.

Step 2. Calculation.

Evaluating $p\left(4\right)\text{and}p\left(-2\right)$ by plugging $x=4\text{and}x=-2$ in given function $p\left(x\right)=\frac{1}{8}{x}^3-\frac{1}{4}{x}^2-\frac{1}{2}x+5$.

$\begin{array}{c}p\left(x\right)=\frac{1}{8}{x}^3-\frac{1}{4}{x}^2-\frac{1}{2}x+5....\left(\text{Given}\right)\\ p\left(4\right)=\frac{1}{8}{\left(4\right)}^3-\frac{1}{4}{\left(4\right)}^2-\frac{1}{2}\left(4\right)+5\\ =7\\ \text{Now,}p\left(-2\right)=\frac{1}{8}{\left(-2\right)}^3-\frac{1}{4}{\left(-2\right)}^2-\frac{1}{2}\left(-2\right)+5\\ =4\end{array}$

Step 3. Conclusion.

The value of $p\left(4\right)=7\text{and}p\left(-2\right)=4$.