1) f(x)-g(x) = x² - (-2x+3) = x² + 2x - 3
et (x+3) (x-1) = x² - x + 3x - 3 = x² + 2x - 3 = f(x) - g(x)
2)
f(x) > g(x)
f(x) - g(x) > 0
signe de (x+3) (x-1) ?
x -inf -3 1 +inf
x+3 - 0 + +
x-1 - - 0 +
final + 0 - 0 +
f(x) - g(x) > 0 et donc f(x) > g(x) sur ] - inf ; -3 [ U ] 1 ; + inf[
