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Question Number 136270 by EDWIN88 last updated on 20/Mar/21

$$\mathrm{What}\mathrm{is}\mathrm{range}\mathrm{of}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=(\mathrm{x}+1)(\mathrm{x}+2)(\mathrm{x}+3)(\mathrm{x}+4)+1$$$$\mathrm{where}\mathrm{x}\in [-1,1]$$

Answered by mr W last updated on 20/Mar/21

$$f\left(x\right)=(x^2+5x+4)(x^2+5x+6)+1$$$$f\left(x\right)=(x^2+5x+4)(x^2+5x+4+2)+1$$$$f\left(x\right)={(x^2+5x+4)}^2+2(x^2+5x+4)+1$$$$f\left(x\right)={(x^2+5x+5)}^2$$$$f\left(x\right)={[{(x+\frac{5}{2})}^2-\frac{5}{4}]}^2⩾0$$$${(x+\frac{5}{2})}^2-\frac{5}{4}=0$$$$x=-\frac{5+\sqrt{5}}{2}\approx -3.618,-\frac{5-\sqrt{5}}{2}\approx -1.382$$$$i.e.in(-\mathrm{\infty },-3.618]f\left(x\right)isdecreasing$$$$in[-1.382,+\mathrm{\infty })f\left(x\right)isincreasing$$$$\Rightarrow in[-1,1]:$$$$min.f\left(x\right)=f(-1)=1$$$$max.f\left(x\right)=f\left(1\right)=2\times 3\times 4\times 5+1=121$$$$i.e.rangeoff\left(x\right)in[-1,1]is[1,121].$$

Commented by mr W last updated on 20/Mar/21

Commented by EDWIN88 last updated on 20/Mar/21

$$\mathrm{great}..$$

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