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Question Number 184819 by cortano1 last updated on 12/Jan/23
For 0≤x≤1 , maximum value of f(x)=x(√(1−x+(√(1−x)))) is __
For0⩽x⩽1,maximumvalueoff(x)=x1−x+1−xis__
Answered by Frix last updated on 12/Jan/23
f′(x)=0 −((5x−4+2(3x−2)(√(1−x)))/(4(√(1−x))(√(1−x+(√(1−x))))))=0 2(3x−2)(√(1−x))=4−5x Squaring and transforming x(x^2 −((59)/(36))x+(2/3))=0 x=0 [obviously wrong as f′(0)=(√2)] x=(8/9) [also wrong as f′((8/9))=−1] x=(3/4) [f′((3/4))=0] f((3/4))=((3(√3))/8)
f′(x)=0−5x−4+2(3x−2)1−x41−x1−x+1−x=02(3x−2)1−x=4−5xSquaringandtransformingx(x2−5936x+23)=0x=0[obviouslywrongasf′(0)=2]x=89[alsowrongasf′(89)=−1]x=34[f′(34)=0]f(34)=338
Answered by Frix last updated on 12/Jan/23
Let t=(√(1−x))≥0 ⇔ x=1−t^2 f(t)=(1−t^2 )(√(t^2 +t)) f′(t)=0 −((6t^3 +5t^2 −2t−1)/(2(√(t^2 +t))))=0 t=−1 [wrong] t=−(1/3) [wrong] t=(1/2) ⇒ x=(3/4)
Lett=1−x⩾0⇔x=1−t2f(t)=(1−t2)t2+tf′(t)=0−6t3+5t2−2t−12t2+t=0t=−1[wrong]t=−13[wrong]t=12⇒x=34

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