Answer
$P\left( \dfrac{1}{2} \right)=\dfrac{95}{32}$
Work Step by Step
Substituting $x$ with $
\dfrac{1}{2}
$ in $
P(x)=x^5+x^4-x^3+3
$, then
\begin{array}{l}
P\left( \dfrac{1}{2} \right)=\left( \dfrac{1}{2} \right)^5+\left( \dfrac{1}{2} \right)^4-\left( \dfrac{1}{2} \right)^3+3
\\\\
P\left( \dfrac{1}{2} \right)=\dfrac{1}{32}+\dfrac{1}{16}-\dfrac{1}{8}+3
\\\\
P\left( \dfrac{1}{2} \right)=\dfrac{95}{32}
.\end{array}
