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 ]](https://www.tinkutara.com/question/Q136270.png)
$$\mathrm{What}\:\mathrm{is}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{4}\right)+\mathrm{1} \\ $$$$\mathrm{where}\:\mathrm{x}\:\in\:\left[\:−\mathrm{1},\:\mathrm{1}\:\right] \\ $$
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].](https://www.tinkutara.com/question/Q136274.png)
$${f}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}\right)+\mathrm{1} \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{4}+\mathrm{2}\right)+\mathrm{1} \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{4}\right)^{\mathrm{2}} +\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{4}\right)+\mathrm{1} \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{5}\right)^{\mathrm{2}} \\ $$$${f}\left({x}\right)=\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right]^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{2}}\approx−\mathrm{3}.\mathrm{618},\:−\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}}\approx−\mathrm{1}.\mathrm{382} \\ $$$${i}.{e}.\:{in}\:\left(−\infty,−\mathrm{3}.\mathrm{618}\right]\:{f}\left({x}\right)\:{is}\:{decreasing} \\ $$$${in}\:\left[−\mathrm{1}.\mathrm{382},+\infty\right)\:{f}\left({x}\right)\:{is}\:{increasing} \\ $$$$\Rightarrow\:{in}\:\left[−\mathrm{1},\mathrm{1}\right]: \\ $$$${min}.\:{f}\left({x}\right)={f}\left(−\mathrm{1}\right)=\mathrm{1} \\ $$$${max}.\:{f}\left({x}\right)={f}\left(\mathrm{1}\right)=\mathrm{2}×\mathrm{3}×\mathrm{4}×\mathrm{5}+\mathrm{1}=\mathrm{121} \\ $$$${i}.{e}.\:{range}\:{of}\:{f}\left({x}\right)\:{in}\:\left[−\mathrm{1},\mathrm{1}\right]\:{is}\:\left[\mathrm{1},\mathrm{121}\right]. \\ $$
Commented by mr W last updated on 20/Mar/21
Commented by EDWIN88 last updated on 20/Mar/21
$$\mathrm{great}.. \\ $$