Answer
$P\left( \dfrac{1}{2} \right)=-\dfrac{17}{4}$
Work Step by Step
Substituting $x$ with $
\dfrac{1}{2}
$ in $
P(x)=4x^4-2x^3+x^2-x-4
$, then
\begin{array}{l}
P\left( \dfrac{1}{2} \right)=4\left( \dfrac{1}{2} \right)^4-2\left( \dfrac{1}{2} \right)^3+\left( \dfrac{1}{2} \right)^2-\left( \dfrac{1}{2} \right)-4
\\\\
P\left( \dfrac{1}{2} \right)=4\left( \dfrac{1}{16} \right)-2\left( \dfrac{1}{8} \right)+\left( \dfrac{1}{4} \right)-\left( \dfrac{1}{2} \right)-4
\\\\
P\left( \dfrac{1}{2} \right)=\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{2}-4
\\\\
P\left( \dfrac{1}{2} \right)=-\dfrac{17}{4}
.\end{array}
