To calculate "For which values of #m# #f# has no etreme points" first we have to calculate the derivative:
#f(x)=(m-x)/(x^2-3x+2)#
#f'(x)=((m-x)'(x^2-3x+2)-(m-x)(x^2-3x+2)')/(x^2-3x+2)^2#
#f'(x)=(-x^2+3x-2-(m-x)(2x-3))/(x^2-3x+2)^2#
#f'(x)=(-x^2+3x-2-2mx+3m+2x^2-3x)/(x^2-3x+2)^2#
#f'(x)=(x^2-2mx+3m-2)/(x^2-3x+2)^2#
The function #f(x)# has extreme values if #f'(x)=0# for some values of #x#, so to calculate when #f(x)# has no extreme points we have to find values of #m# for which the numerator of #f'x)# has no zeros.
The equation #x^2-2mx+3m-2=0# has no solutions if its discriminant is negative.
The coefficients are:
#a=1#, #b=-2m#, #c=3m-2#
The discriminant is:
#Delta=(-2m)^2-4*(3m-2)=4(m^2-3m+2)#
Now we have to solve the inequality #m^2-3m+2<0#
#Delta=(-3)^2-4*1*2=9-8=1#
#m_1=(3-1)/2=1#, #m_2=(3+1)/2=2#
From the graph:
graph{x^2-3x+2 [-0.2, 4, -2.19, 2.194]}
we see that the discriminant is negative (i.e. the derivative has no zeros) for #m in (1;2)#
Answer: The function #f(x)# has no extreme points for #m in (1;2)#.
