Question

DAU COROANA!URGENT!!

Answer 1

a) $f'(x)=e^x-1$



b) $f'(x)=0\implies e^x=1\implies x=0$

$x < 0\implies e^x < 1\implies f'(x) < 0$

$\implies\boxed{ f \text{ descrescatoare pe }(-\infty, 0)}$

$\text{Analog }\boxed{f\text{ crescatoare pe }(0,\infty)}$

$\text{iar }\boxed{x=0}\text{ este punct de minim}$



c) $e^{x^2}+e^x\ge x^2+x+2\iff e^{x^2}+e^x-x^2-x-2\ge0$

se observa ca $f(x^2)+f(x)=e^{x^2}-x^2-1+e^x-x-1=e^{x^2}+e^x-x^2-x-2$

dar stim de mai devreme ca $x=0$ este punct de minim pentru $f$

inseamna ca $f(x)\ge f(0) \; \forall x\in\mathbb{R}$

dar $f(0)=e^0-0-1=1-1=0$

deci $f(x)\ge 0 \; \forall x\in\mathbb{R}$

$\implies f(x^2)+f(x)\ge0\ \forall x\in\mathbb{R}$

$\implies e^{x^2}+e^x-x^2-x-2\ge0\ \forall x\in\mathbb{R}$

$\implies\boxed{e^{x^2}+e^x\ge x^2+x+2\ \forall x\in\mathbb{R}}$



d) $f''(x)=e^x\ge0\ \forall x\in\mathbb{R}\implies \boxed{f \text{ convexa}}$


e) limita aceea este tocmai definitia derivatei lui f in punctul x=1, deci

$\displaystyle \lim_{x\to1}\cfrac{f(x)-f(1)}{x-1}=f'(1)=e^1-1=\boxed{e-1}$



f) ecuatia tangentei la grafic in punctul A(0, 0) este:

$\displaystyle y-f(0)=f'(0)(x-0)$

dar $f(0)=f'(0)=0$, deci ecuatia tangentei este

$\boxed{y=0}$