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Find all the values of the parameter a for which the point of minimum of the function f(x)=1+a2x-x3 satisfies the inequality ((x2+x+2)/(x2+5x+6))>0
Dear harshit
((x2+x+2)/(x2+5x+6))>0
x2+x+2 is always positive
so x2+5x+6 should be >0
which give
x<-3 and x>-2
Now
f(x)=1+a2x-x3
for minimum value differentiate
f'(x)=a2-3x2
f''(x)=-6x
equate f'(x) =0
fives x=+-a/√3
case 1
a>0
x=-a/√3 gives minimum value
so -a/√3 <-3 and -a/√3 >-2
a>3√3 and a<2√3
so 0<a<2√3 and a>3√3
case 2
a<0
x=a/√3 gives minimum value
so a/√3 <-3 and a/√3 >-2
a<-3√3 and a>-2√3
so a<-3√3 and -2√3<a<0
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