Question

Fie o functie f:R⇒R, f(x) = x+ $e^{x}$. Demonstrati ca, daca f(x)≥mx+1, ∨x∈R atunci m=2.

Answer 1

$f(x) = x+e^x\\ \\ f(x)\geq mx+1 \\ x+e^x \geq mx+1 \\ e^x+x-mx \geq 1 \\ e^x-(m-1)x \geq 1 \\ \\ g(x) = e^x-(m-1)x \\ g'(x) = e^x -(m-1) = 0\\ \Rightarrow e^x = m-1 \Rightarrow x_0= \ln(m-1)\\ \\ \Rightarrow \Big(x_0, g(x_0)\Big)\to \text{punct de minim.} \\ \\ g(x_0) = 1 \Rightarrow e^{\ln(m-1)}-(m-1)\cdot \ln(m-1)= 1 \\\\\Rightarrow m-1 - (m-1)\cdot \ln(m-1) = 1 \\ \Rightarrow 1 - \ln(m-1) = \dfrac{1}{m-1} \\\\\Rightarrow \ln(m-1) = \dfrac{m-2}{m-1}\\ \\ \Rightarrow \boxed{m= 2}$