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Question Number 121278 by Ar Brandon last updated on 06/Nov/20
The domaine of the function  f(x)=((√(log_(0.5) (x^2 −7x+13))))^(−1)  is;
Thedomaineofthefunctionf(x)=(log0.5(x27x+13))1is;
Answered by TANMAY PANACEA last updated on 06/Nov/20
x^2 −7x+13  (x^2 −2×x×(7/2)+((49)/4)+13−((49)/4))  (x−(7/2))^2 +(3/4)>0  so  g(x)=x^2 −7x+13>0  f(x)=(1/( (√(log_(0.5) (x^2 −7x+13)))))  note (x^2 −7x+13) can not be equsls to 1    if x^2 −7x+13=1  f(x)=(1/( (√(log_(0.5) 1))))=(1/0)=unddfined  so 1>(x^2 −7x+13)>0  x^2 −7x+13<1  x^2 −7x+12<0  (x−3)(x−4)<0  when x>4    (x−3)(x−4)>0 not feasibld  when x<3  (x−3)(x−4)>0 not feasible  when 4>x>3 (x−3)(x−4)<0 feasible  so   domain ...4>x>3
x27x+13(x22×x×72+494+13494)(x72)2+34>0sog(x)=x27x+13>0f(x)=1log0.5(x27x+13)note(x27x+13)cannotbeequslsto1ifx27x+13=1f(x)=1log0.51=10=unddfinedso1>(x27x+13)>0x27x+13<1x27x+12<0(x3)(x4)<0whenx>4(x3)(x4)>0notfeasibldwhenx<3(x3)(x4)>0notfeasiblewhen4>x>3(x3)(x4)<0feasiblesodomain4>x>3
Commented by TANMAY PANACEA last updated on 06/Nov/20
Commented by Ar Brandon last updated on 06/Nov/20
Thanks Sir
Commented by TANMAY PANACEA last updated on 06/Nov/20
most selcome sir
mostselcomesir

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