R Richard AndrÃ©-Jeannin Oct 2007 121 1 France Jan 21, 2008 #2 f(x)=sqrt(u(x)), where u(x)=x^3/(x-2) The function f is defined for x>2 or x<=0 f '(x)=u '/(2sqrt(u)) u '(x)=2xÂ²(x-3)/(x-2)Â² The function f ' is defined for x>2 or x<0. f '(x)=0 if x=3 f '(x)>0 for x>3 and f '(x)<0 for x<3

f(x)=sqrt(u(x)), where u(x)=x^3/(x-2) The function f is defined for x>2 or x<=0 f '(x)=u '/(2sqrt(u)) u '(x)=2xÂ²(x-3)/(x-2)Â² The function f ' is defined for x>2 or x<0. f '(x)=0 if x=3 f '(x)>0 for x>3 and f '(x)<0 for x<3

M morph Nov 2007 13 0 Jan 22, 2008 #3 Richard AndrÃ©-Jeannin said: f(x)=sqrt(u(x)), where u(x)=x^3/(x-2) The function f is defined for x>2 or x<=0 f '(x)=u '/(2sqrt(u)) u '(x)=2xÂ²(x-3)/(x-2)Â² The function f ' is defined for x>2 or x<0. f '(x)=0 if x=3 f '(x)>0 for x>3 and f '(x)<0 for x<3 Click to expand... thanks...I see I can solve it this way! but still wondering why tabular way isn't working...(where's my mistake?)

Richard AndrÃ©-Jeannin said: f(x)=sqrt(u(x)), where u(x)=x^3/(x-2) The function f is defined for x>2 or x<=0 f '(x)=u '/(2sqrt(u)) u '(x)=2xÂ²(x-3)/(x-2)Â² The function f ' is defined for x>2 or x<0. f '(x)=0 if x=3 f '(x)>0 for x>3 and f '(x)<0 for x<3 Click to expand... thanks...I see I can solve it this way! but still wondering why tabular way isn't working...(where's my mistake?)