f(x)=3x+e^x
f(x)≥4x+1
3x+e^x≥4x+1
3x+e^x-4x-1≥0
e^x-x-1≥0
Formam o noua functie g: R->R, g(x)=e^x-x-1
g'(x)=(e^x-x-1)'=e^x-1
g'(x)=0 => e^x-1=0 => e^x=1 => x=0
Facem tabelul de semn:
x |-∞_________________0_________________∞
g'|--------------------------------0++++++++++++++++
g | descrecatoare g(0) crecscatoare
g'(-1)=e^-1-1=1/e-1=(1-e)/e<0
g'(1)=e^1-1=e-1>0
x=0 funct de minim =>
g(x)≥g(0)
g(0)=e^0-0-1=1-1=0
Deci inegalitatea va fi:
g(x)≥0
e^x-x-1≥0
e^x≥x+1
Adunand 3x in ambii membrii avem:
e^x+3x≥4x+1
f(x)≥4x+1
