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Question Number 136270 by EDWIN88 last updated on 20/Mar/21
What is range of function f(x)=(x+1)(x+2)(x+3)(x+4)+1  where x ∈ [ −1, 1 ]
Whatisrangeoffunctionf(x)=(x+1)(x+2)(x+3)(x+4)+1wherex[1,1]
Answered by mr W last updated on 20/Mar/21
f(x)=(x^2 +5x+4)(x^2 +5x+6)+1  f(x)=(x^2 +5x+4)(x^2 +5x+4+2)+1  f(x)=(x^2 +5x+4)^2 +2(x^2 +5x+4)+1  f(x)=(x^2 +5x+5)^2   f(x)=[(x+(5/2))^2 −(5/4)]^2 ≥0  (x+(5/2))^2 −(5/4)=0  x=−((5+(√5))/2)≈−3.618, −((5−(√5))/2)≈−1.382  i.e. in (−∞,−3.618] f(x) is decreasing  in [−1.382,+∞) f(x) is increasing  ⇒ in [−1,1]:  min. f(x)=f(−1)=1  max. f(x)=f(1)=2×3×4×5+1=121  i.e. range of f(x) in [−1,1] is [1,121].
f(x)=(x2+5x+4)(x2+5x+6)+1f(x)=(x2+5x+4)(x2+5x+4+2)+1f(x)=(x2+5x+4)2+2(x2+5x+4)+1f(x)=(x2+5x+5)2f(x)=[(x+52)254]20(x+52)254=0x=5+523.618,5521.382i.e.in(,3.618]f(x)isdecreasingin[1.382,+)f(x)isincreasingin[1,1]:min.f(x)=f(1)=1max.f(x)=f(1)=2×3×4×5+1=121i.e.rangeoff(x)in[1,1]is[1,121].
Commented by mr W last updated on 20/Mar/21
Commented by EDWIN88 last updated on 20/Mar/21
great..
great..

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