Question

Aratati ca f(x)=x(x²-3x)+3

Answer 1

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. $\lim_{x \to \infty} \frac{x^{3} -3x+3-x^{3} }{x+1} = \lim_{x \to \infty} \frac{ -3x+3 }{x+1}=-3$ (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. $\int\limits {x^{4} } \, dx =\frac{x^{5} }{5}$| de la -1 la 1= $\frac{1}{5} +\frac{1}{5} =\frac{2}{5}$

b. $\int\limits^e_1 {xlnx} \, dx =lnx\cdot \frac{x^{2} }{2} |\limits^e_1-\int\limits^e_1 {\frac{x}{2} } \, dx =lnx\cdot \frac{x^{2} }{2} |\limits^e_1-\frac{x^{2} }{4} |\limits^e_1=\frac{e^{2} }{2} -\frac{e^{2} }{4} +\frac{1}{4} =\frac{e^{2}+1 }{4}$

f=lnx           f'=$\frac{1}{x}$

g'=x             g=$\frac{x^{2} }{2}$

c.

$\int\limits^a_0 {x^{4} +x+e^{x} } \, dx =\frac{x^{5} }{5} |\limits^a_0+\frac{x^{2} }{2} |\limits^a_0+e^{x} |\limits^a_0$

$\frac{a^{5} }{5} +\frac{a^{2} }{2} +e^{a} -1=\frac{5a^{2} +54}{10} +e^{a}\\\\ \frac{2a^{5}+5a^{2} -10 }{10} =\frac{5a^{2} +54}{10}$

2a⁵-10=54

2a⁵=64

a⁵=32

a=2