We have #f:RR->RR,f(x)=x^2-(m-1)x+3m-4,m inRR#.How to find the values of #m# for which #f# has root in #(0,1)# and #f(x)>=0,forall x in(0,1)#?

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Answer 1 · 2017-05-20T20:32:27

See below.

Given that #f# has a root in #(0,1)# and that #f(x) >= 0#, the root must have multiplicity #2#,

So, we must have #f(x) = (x-a)^2# for some #a# in #(0,1)#

Set

#x^2-(m-1)x+3m-4 = (x-a)^2# and solve for #m#.

#x^2-(m-1)x+3m-4 = x^2-2ax+a^2#

The coefficients of each term must be equal, so

#2a=m-1# and #a^2=3m-4#

So, we need

#a=(m-1)/2# and, therefore,

#((m-1)/2)^2=3m-4#.

This leads to #m^2-14m+17 = 0#, and finally

#m = 7+-4sqrt2#