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Question-203063




Question Number 203063 by LowLevelLump last updated on 09/Jan/24
Answered by MM42 last updated on 09/Jan/24
f′=e^x −a=0⇒α=lna  ⇒minf=a−alna   g′=a−(1/x)=0⇒β=(1/a)  ⇒ming=1+lna  ⇒minf=ming⇒a−alna=1+lna  ⇒a−1=(a+1)lna⇒a=1 ✓  ⇒f=e^x −x   &   g=x−lnx
f=exa=0α=lnaminf=aalnag=a1x=0β=1aming=1+lnaminf=mingaalna=1+lnaa1=(a+1)lnaa=1f=exx&g=xlnx
Commented by MM42 last updated on 09/Jan/24

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