Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 56900 by bshahid010@gmail.com last updated on 26/Mar/19

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Mar/19

f(x)=∣(x−a)^3 ∣+∣(x−b)^3 ∣ =(x−a)^3 +(x−b)^3 when x>b =(b−a)^3 when x=b =(x−a)^3 −(x−b)^3 when b>x>a =−(a−b)^3 when x=a =(b−a)^3 =−(x−a)^3 −(x−b)^3 when x<a when x>b f(x)=(x−a)^3 +(x−b)^3 (df/dx)=3(x−a)^2 +3(x−b)^2 (df/dx)≠0 as x>b but for max/min (df/dx)=0 (d^2 f/dx^2 )=6(x−a)+6(x−b) so no max/min value at x>b when b>x>a f(x)=(x−a)^3 −(x−b)^3 (df/dx)=3(x−a)^2 −3(x−b)^2 (df/dx)=0 at x=((a+b)/2) (d^2 f/dx^2 )=6(x−a)−6(x−b)=6(b−a)>0 so at x=((a+b)/2) i,e f(((a+b)/2))=minimum f(((a+b)/2))=(((a+b)/2)−a)^3 −(((a+b)/2)−b)^3 =(((b−a)/2))^3 −(((a−b)/2))^3 =2×(((b−a)/2))^3 ←this is the answer (minimum value) when x<a f(x)=−(x−a)^3 −(x−b)^3 (df/dx)=−3[(x−a)^2 +(x−b)^2 ] but (df/dx)≠0 so no min/max when x<a

f(x)=∣(x−a)3∣+∣(x−b)3∣=(x−a)3+(x−b)3whenx>b=(b−a)3whenx=b=(x−a)3−(x−b)3whenb>x>a=−(a−b)3whenx=a=(b−a)3=−(x−a)3−(x−b)3whenx<awhenx>bf(x)=(x−a)3+(x−b)3dfdx=3(x−a)2+3(x−b)2dfdx≠0asx>bbutformax/mindfdx=0d2fdx2=6(x−a)+6(x−b)sonomax/minvalueatx>bwhenb>x>af(x)=(x−a)3−(x−b)3dfdx=3(x−a)2−3(x−b)2dfdx=0atx=a+b2d2fdx2=6(x−a)−6(x−b)=6(b−a)>0soatx=a+b2i,ef(a+b2)=minimumf(a+b2)=(a+b2−a)3−(a+b2−b)3=(b−a2)3−(a−b2)3=2×(b−a2)3←thisistheanswer(minimumvalue)whenx<af(x)=−(x−a)3−(x−b)3dfdx=−3[(x−a)2+(x−b)2]butdfdx≠0sonomin/maxwhenx<a

Terms of Service

Privacy Policy

Contact: info@tinkutara.com