Răspuns:
Explicație pas cu pas:
Ex1
f(x)=(x-1)²eˣ
a) f'(x)=[(x-1)²]'eˣ+(x-1)²(eˣ)'=2(x-1)·(x-1)'eˣ+(x-1)²eˣ=eˣ(x-1)(2x-2+x-1)=eˣ(x-1)(3x-3)=3eˣ(x-1)²
b) f'(x)=0, 3eˣ(x-1)²=0, x=1 punct critic.
prntru x∈(-∞; 1), f'(x)>0; pentru x∈(1;+∞) , f'(x)>0, deci c=1 nu este punct de extrem ci punct de inflexiune.
Deci nr. de puncte de extrem este 0.
c) [tex]\lim_{x \to \infty} x*[\frac{f'(x)}{f(x)}-1]= \lim_{x \to \infty} x*[\frac{3*e^{x}*(x-1)^{2}}{(x-1)^{2}*e^{x}}-1]= \lim_{x \to \infty} (x*2)=\infty[/tex]
ex2
[tex]\int\limits^1_0 {f(x)e^{-x}} \, dx= \int\limits^1_0 {(x-1)^{2}e^{x}e^{-x}} \, dx= \int\limits^1_0 {f(x)e^{-x}} \, dx= \int\limits^1_0 {(x-1)^{2}e^{x}e^{-x}} \, dx= \int\limits^1_0 {(x-1)^{2}} \, dx=\frac{(x-1)^{3}}{3}|^{1}_{0}=\frac{(1-1)^{3}}{3} - \frac{(0-1)^{3}}{3}=0+\frac{1}{3}= \frac{1}{3}.[/tex]
