Question

Va rog dau coroana . ​

Answer 1

Explicație pas cu pas:

a) asimptotă oblică

$lim_{x\rightarrow + \infty }\left(\frac{f(x)}{x} \right) = lim_{x\rightarrow + \infty }\left(\frac{ {x}^{2} + x + 1}{x(x + 1)} \right) \\ = lim_{x\rightarrow + \infty }\left(\frac{ {x}^{2} + x + 1}{ {x}^{2} + x} \right) = lim_{x\rightarrow + \infty }\left(\frac{ {x}^{2} (1 + \frac{1}{x} + \frac{1}{ {x}^{2} } )}{ {x}^{2}(1 + \frac{1}{x})} \right) \\ = lim_{x\rightarrow + \infty }\left(\frac{1 + \frac{1}{x} + \frac{1}{ {x}^{2} } }{1 + \frac{1}{x}} \right) = \frac{1 + 0 + 0}{1 + 0 + 0} = \frac{1}{1} = 1$

$lim_{x\rightarrow + \infty }\left(f(x) - 1\cdot x \right) = lim_{x\rightarrow + \infty }\left(\frac{ {x}^{2} + x + 1}{x + 1} - x \right) \\ = lim_{x\rightarrow + \infty }\left(\frac{ {x}^{2} + x + 1 - x(x + 1)}{x + 1}\right) \\ = lim_{x\rightarrow + \infty }\left(\frac{ {x}^{2} + x + 1 - {x}^{2} - x}{x + 1}\right) \\ = lim_{x\rightarrow + \infty }\left(\frac{1}{x + 1}\right) = \frac{1}{ + \infty } = 0$

→ ecuația asimptotei oblice a graficului funcției f către +∞:

$y = 1\cdot x + 0 = > y = x$

b) derivata funcției f:

$f^{\prime}(x) = \left(\frac{ {x}^{2} + x + 1}{x + 1}\right)^{\prime} \\ = \frac{\left({x}^{2} + x + 1 \right)^{\prime}(x + 1) - ({x}^{2} + x + 1)\left(x + 1 \right)^{\prime}}{ {\left(x + 1\right)}^{2} } \\ = \frac{(2x + 1)(x + 1) - ({x}^{2} + x + 1)}{ {\left(x + 1\right)}^{2} } \\ = \frac{2 {x}^{2} + 3x + 1 - {x}^{2} - x - 1}{ {\left(x + 1\right)}^{2} } \\ = \frac{{x}^{2} + 2x}{ {\left(x + 1\right)}^{2} } = \frac{x(x + 2)}{ {\left(x + 1\right)}^{2} }$

c) punctele de extrem ale funcției:

$f^{\prime}(x) = 0 < = > \frac{x(x + 2)}{ {\left(x + 1\right)}^{2} } = 0 \\ x(x + 2) = 0 = > \\ x = - 2 \\ x = 0$

$f(-2) = -3 =>maxim \left(-2 ; -3\right)$

$f(0) = 1 =>minin \left(0 ; 1\right)$