Răspuns:
Explicație pas cu pas:
f(x)=x(x²-3)+3
a. f'(x)=x²-3+x×2x=3x²-3=3(x²-1)=3(x-1)(x+1)
b. [tex]\lim_{x \to \infty} \frac{x^{3} -3x+3-x^{3} }{x+1} = \lim_{x \to \infty} \frac{ -3x+3 }{x+1}=-3[/tex] (acelasi grad)
c. y-f(x₀)=f'(x₀)(x-x₀)
f(0)=3
f'(0)=-3
ecuatia tangentei in 0
y-3=-3(x-0)
y-3=-3x
y=-3x+3
2. f(x)=x⁴+x+eˣ
a. [tex]\int\limits {x^{4} } \, dx =\frac{x^{5} }{5}[/tex]| de la -1 la 1= [tex]\frac{1}{5} +\frac{1}{5} =\frac{2}{5}[/tex]
b. [tex]\int\limits^e_1 {xlnx} \, dx =lnx\cdot \frac{x^{2} }{2} |\limits^e_1-\int\limits^e_1 {\frac{x}{2} } \, dx =lnx\cdot \frac{x^{2} }{2} |\limits^e_1-\frac{x^{2} }{4} |\limits^e_1=\frac{e^{2} }{2} -\frac{e^{2} }{4} +\frac{1}{4} =\frac{e^{2}+1 }{4}[/tex]
f=lnx f'=[tex]\frac{1}{x}[/tex]
g'=x g=[tex]\frac{x^{2} }{2}[/tex]
c.
[tex]\int\limits^a_0 {x^{4} +x+e^{x} } \, dx =\frac{x^{5} }{5} |\limits^a_0+\frac{x^{2} }{2} |\limits^a_0+e^{x} |\limits^a_0[/tex]
[tex]\frac{a^{5} }{5} +\frac{a^{2} }{2} +e^{a} -1=\frac{5a^{2} +54}{10} +e^{a}\\\\ \frac{2a^{5}+5a^{2} -10 }{10} =\frac{5a^{2} +54}{10}[/tex]
2a⁵-10=54
2a⁵=64
a⁵=32
a=2
