8x^{5}-3x^{4}+7x^{3}-2x^{2}+2x-1 the "x" is -1/2 as fraction
Given:
$\displaystyle f(x)=8x^5-3x^4+7x^3-2x^2+2x-1$
I take it you are to find:
$\displaystyle f\left(-\frac{1}{2} \right)=8\left(-\frac{1}{2} \right)^5-3\left(-\frac{1}{2} \right)^4+7\left(-\frac{1}{2} \right)^3-2\left(-\frac{1}{2} \right)^2+2\left(-\frac{1}{2} \right)-1=$
$\displaystyle -\frac{8}{32}-\frac{3}{16}-\frac{7}{8}-\frac{2}{4}-\frac{2}{2}-1=$
Can you finish?