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Question Number 210248 by universe last updated on 04/Aug/24
Answered by mr W last updated on 04/Aug/24
f″(x)>0⇒f′(x)isstrictlyincreasing.case1:f′(x)<0⇒f(x)isdecreasingx+f′(x)<x⇒f(x+f′(x))>f(x)case2:f′(x)=0⇒f(x+f′(x))=f(x)case3:f′(x)>0⇒f(x)isincreasingx+f′(x)>x⇒f(x+f′(x))>f(x)summary:f(x+f′(x))⩾f(x)⇒answer(b)istrue
Commented by universe last updated on 04/Aug/24
thanlyousomuchsir
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