$解：將?x=0.5?代入得?\frac 12×0.5^2+2×0.5-1=\frac {1}{8}＞0?$
$∴?x=0.5?在該交點的右側(cè)$
$將?x=0.4?代入得?\frac 12×0.4^2+2×0.4-1=-0.12＜0?$
$∴?x=0.4?在該交點的左側(cè)$
$將?x=0.45?代入得?\frac 12×0.45^2+2×0.45-1=0.00125＞0?$
$∴?x=0.45?在該交點的右側(cè)$
$∴該交點的橫坐標在?0.4?和?0.45?之間$
$則正根的近似值為?0.4?$
