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Question Number 12697 by @ANTARES_VY last updated on 29/Apr/17

y=∣x−2∣+2x−3x^2 find the largest values of the function.

y=∣x−2∣+2x−3x2findthelargestvaluesofthefunction.

Answered by mrW1 last updated on 29/Apr/17

if x≥2: y=x−2+2x−3x^2 =−3(x^2 −x+(1/4))−2+(3/4)=−3(x−(1/2))^2 −(5/4) y_(max) =−(5/4) at x=(1/2)≱2 ! if x<2: y=−x+2+2x−3x^2 =−3(x^2 −(1/3)x+(1/(36)))+2+(1/(12))=−3(x−(1/6))^2 +((25)/(12)) y_(max) =((25)/(12)) at x=(1/6)<2 ok! max. value of function =((25)/(12)) at x=(1/6)

ifx⩾2:y=x−2+2x−3x2=−3(x2−x+14)−2+34=−3(x−12)2−54ymax=−54atx=12≱2!ifx<2:y=−x+2+2x−3x2=−3(x2−13x+136)+2+112=−3(x−16)2+2512ymax=2512atx=16<2ok!max.valueoffunction=2512atx=16

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