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Let-f-x-x-2-x-4-2x-6-for-2-x-8-The-sum-of-the-largest-and-smallest-values-of-f-x-is-

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Question Number 110359 by Aina Samuel Temidayo last updated on 28/Aug/20
Let f(x)=∣x−2∣+∣x−4∣−∣2x−6∣, for 2≤x≤8. The sum of the largest and smallest values of f(x) is
$$\mathrm{Let} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mid\mathrm{x}−\mathrm{2}\mid+\mid\mathrm{x}−\mathrm{4}\mid−\mid\mathrm{2x}−\mathrm{6}\mid, \\ $$$$\mathrm{for}\:\mathrm{2}\leqslant\mathrm{x}\leqslant\mathrm{8}.\:\mathrm{The}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{largest}\:\mathrm{and} \\ $$$$\mathrm{smallest}\:\mathrm{values}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is} \\ $$
Answered by bobhans last updated on 28/Aug/20
∣x−2∣ = { ((x−2 ; x ≥2)),((−x+2; x< 2)) :} ∣x−4∣ = { ((x−4 ; x ≥ 4)),((−x+4 ; x<4)) :} ∣2x−6∣ = { ((2x−6 ; x≥3)),((−2x+6 ; x<3)) :} given 2≤ x≤ 8 →2≤x<3 ∪ 3≤x<4 ∪ 4≤x≤8 case(1) 2≤x<3→f(x)=x−2−(x−4)+2x−6 f(x)=2x−4→ { ((f(2)=0)),((f(3)=2)) :} case(2)3≤x<4→f(x)=x−2−(x−4)−(2x−6) f(x)=−2x+8 → { ((f(3)=2)),((f(4)=0)) :} case(3)4≤x≤8→f(x)=x−2+x−4−(2x−6) f(x)=0 min value of f(x) = 0 & max value of f(x) = 2 Σ(min, max) = 2
$$\mid{x}−\mathrm{2}\mid\:=\:\begin{cases}{{x}−\mathrm{2}\:;\:{x}\:\geqslant\mathrm{2}}\\{−{x}+\mathrm{2};\:{x}<\:\mathrm{2}}\end{cases} \\ $$$$\mid{x}−\mathrm{4}\mid\:=\:\begin{cases}{{x}−\mathrm{4}\:;\:{x}\:\geqslant\:\mathrm{4}}\\{−{x}+\mathrm{4}\:;\:{x}<\mathrm{4}}\end{cases} \\ $$$$\mid\mathrm{2}{x}−\mathrm{6}\mid\:=\:\begin{cases}{\mathrm{2}{x}−\mathrm{6}\:;\:{x}\geqslant\mathrm{3}}\\{−\mathrm{2}{x}+\mathrm{6}\:;\:{x}<\mathrm{3}}\end{cases} \\ $$$${given}\:\mathrm{2}\leqslant\:{x}\leqslant\:\mathrm{8}\:\rightarrow\mathrm{2}\leqslant{x}<\mathrm{3}\:\cup\:\mathrm{3}\leqslant{x}<\mathrm{4}\:\cup\:\mathrm{4}\leqslant{x}\leqslant\mathrm{8} \\ $$$${case}\left(\mathrm{1}\right)\:\mathrm{2}\leqslant{x}<\mathrm{3}\rightarrow{f}\left({x}\right)={x}−\mathrm{2}−\left({x}−\mathrm{4}\right)+\mathrm{2}{x}−\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\mathrm{2}{x}−\mathrm{4}\rightarrow\begin{cases}{{f}\left(\mathrm{2}\right)=\mathrm{0}}\\{{f}\left(\mathrm{3}\right)=\mathrm{2}}\end{cases} \\ $$$${case}\left(\mathrm{2}\right)\mathrm{3}\leqslant{x}<\mathrm{4}\rightarrow{f}\left({x}\right)={x}−\mathrm{2}−\left({x}−\mathrm{4}\right)−\left(\mathrm{2}{x}−\mathrm{6}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=−\mathrm{2}{x}+\mathrm{8}\:\rightarrow\begin{cases}{{f}\left(\mathrm{3}\right)=\mathrm{2}}\\{{f}\left(\mathrm{4}\right)=\mathrm{0}}\end{cases} \\ $$$${case}\left(\mathrm{3}\right)\mathrm{4}\leqslant{x}\leqslant\mathrm{8}\rightarrow{f}\left({x}\right)={x}−\mathrm{2}+{x}−\mathrm{4}−\left(\mathrm{2}{x}−\mathrm{6}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\mathrm{0}\: \\ $$$${min}\:{value}\:{of}\:{f}\left({x}\right)\:=\:\mathrm{0}\:\&\:{max}\:{value}\:{of} \\ $$$${f}\left({x}\right)\:=\:\mathrm{2} \\ $$$$\Sigma\left({min},\:{max}\right)\:=\:\mathrm{2} \\ $$
Commented by Aina Samuel Temidayo last updated on 28/Aug/20
Wow. Thanks. Please help in the remaining questions I posted.
$$\mathrm{Wow}.\:\mathrm{Thanks}.\:\mathrm{Please}\:\mathrm{help}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{remaining}\:\mathrm{questions}\:\mathrm{I}\:\mathrm{posted}. \\ $$
Commented by Aina Samuel Temidayo last updated on 28/Aug/20
Please how did you get this: given 2≤x≤8→ 2≤x<3 ∪ 3≤x<4 ∪ 4≤x≤8
$$\mathrm{Please}\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{this}:\:\mathrm{given} \\ $$$$\mathrm{2}\leqslant\mathrm{x}\leqslant\mathrm{8}\rightarrow\:\mathrm{2}\leqslant\mathrm{x}<\mathrm{3}\:\cup\:\mathrm{3}\leqslant\mathrm{x}<\mathrm{4}\:\cup\:\mathrm{4}\leqslant\mathrm{x}\leqslant\mathrm{8} \\ $$

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