Chapter 6.3, Problem 43PPS

To determine

To find:

The value of $p\left(-2\right),p\left(8\right)$

Answer to Problem 43PPS

The value of, $p\left(-2\right)=-16,p\left(8\right)=1024$

Explanation of Solution

Given:

  $p\left(x\right)=\frac{1}{4}{x}^4+\frac{1}{2}{x}^3-4x^2$

Concept:

Using the given value of the function .

Calculation:

The function

  $p\left(x\right)=\frac{1}{4}{x}^4+\frac{1}{2}{x}^3-4x^2\mathrm{.........}\left(1\right)$

Putting the value $x=-2,8$ in equation (1)

  $\begin{array}{l}p\left(x\right)=\frac{1}{4}{x}^4+\frac{1}{2}{x}^3-4x^2\\ p\left(-2\right)=\frac{1}{4}{\left(-2\right)}^4+\frac{1}{2}{\left(-2\right)}^3-4{\left(-2\right)}^2\\ p\left(-2\right)=4-4-16\\ p\left(-2\right)=-16\\ p\left(8\right)=\frac{1}{4}{\left(8\right)}^4+\frac{1}{2}{\left(8\right)}^3-4{\left(8\right)}^2\\ p\left(8\right)=1024+256-256\\ p\left(8\right)=1024\end{array}$