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TES Octobre 2010page 1 Ex13p88: "x > 0 ln(x) + 4 a) f (x) = x ! 5 + !# !Df = ]0;1["]1;+#[. x !1 $x ! 1 "x > 0 1 ! x2 b) f (x) = !Df = R+*. + x ln(x) ! # 2x $x ! 0 c) f (x) = ln(3 + x) + x 2 ! x ! 3 + x > 0 !Df = ]$3;+#[. h) f (x) = 2x +1 " x > 0 • # !Df = ]0;1["]1;+#[. ln x $ln x ! 0 1 u = 2x +1 u'= 2 v = ln x v' = x 1 2 ln x ! (2x +1) x f '(x) = ( ln x )2 x !1 x !1 )! > 0 !x % ]1;2[!Df = ]1;2[. 2!x 2!x "x > 2 % 2 e) f (x) = 3ln(x ! 2) ! 2 ln(x !1) ! #" x < !1! ! Df = ]2;+#[. %# $$ou!x >