Question

se considera functia f:R->R, f (x)=e^x-x

Answer 1

$\displaystyle a)~f'(x)=e^x-1. \\ \\ b)~Radacina~derivatei~este~0.~Observam~ca: \\ \\ \bullet  Daca~ x \in (-\infty,0] \Rightarrow f'(x) \le e^0-1=0. \\ \\ \bullet Daca~x \in [0, +\infty) \Rightarrow f'(x) \ge e^0-1=0. \\ \\ Deci~f~este~descrescatoare~pe~(-\infty,0]~si~crescatoare~pe~[0,+ \infty).$

$\displaystyle c)~Am~vazut~anterior~ca~f~este~crescatoare~pe~[0,+ \infty). \\ \\ Deoarece~derivata~are~o~singura~radacina~(deci~un~numar~finit~de \\ \\ radacini),~rezulta~ca~f~este~strict~crescatoare~pe~[0,+\infty). \\ \\ Trebuie~sa~demonstram~ca~\sqrt[100]{e}> \frac{101}{100},~echivalent~cu \\ \\ e^{ \frac{1}{100}}>1+ \frac{1}{100} \Leftrightarrow e^{\frac{1}{100}}- \frac{1}{100}>1 \Leftrightarrow f \left( \frac{1}{100} \right)>f(0),~ceea~ce~este$

$\displaystyle adevarat,~caci~\frac{1}{100}>0,~iar~f~este~strict~crescatoare~pe~[0,+\infty).$