Question Number 203063 by LowLevelLump last updated on 09/Jan/24
Answered by MM42 last updated on 09/Jan/24
$${f}'={e}^{{x}} −{a}=\mathrm{0}\Rightarrow\alpha={lna} \\ $$$$\Rightarrow{minf}={a}−{alna} \\ $$$$\:{g}'={a}−\frac{\mathrm{1}}{{x}}=\mathrm{0}\Rightarrow\beta=\frac{\mathrm{1}}{{a}} \\ $$$$\Rightarrow{ming}=\mathrm{1}+{lna} \\ $$$$\Rightarrow{minf}={ming}\Rightarrow{a}−{alna}=\mathrm{1}+{lna} \\ $$$$\Rightarrow{a}−\mathrm{1}=\left({a}+\mathrm{1}\right){lna}\Rightarrow{a}=\mathrm{1}\:\checkmark \\ $$$$\Rightarrow{f}={e}^{{x}} −{x}\:\:\:\&\:\:\:{g}={x}−{lnx} \\ $$$$ \\ $$
Commented by MM42 last updated on 09/Jan/24
