Question

f:R-R , f(x)={x la a doua -x


x>1 {-x la a doua +x, x>1




Sa se calculeze continuitate funcției f in x0=1


Caluculati f’(0)+f’(2)



Ps: este o singura paranteza sus

Answer 1

Pozele conțin rezolvarea.

Answer 2

$\it f: \mathbb{R} \longrightarrow \mathbb{R} ,\ \ f(x) = \begin{cases} \it x^2-x,\ x \leq1\\ \\ \it -x^2+x,\ x>1\end{cases}\\ \\ \\ \lim\limits_{x \to 1 \ x<1} f(x) = \lim\limits_{x \to 1 \ x<1} (x^2-x) = 0 \ \ \ \ \ \ \ (1)\\ \\ \\ \lim\limits_{x \to 1 \ x>1} f(x) = \lim\limits_{x \to 1 \ x>1} (-x^2+x) = 0 \ \ \ \ \ (2)\\ \\ \\ f(1) = 1^2-1=0 \ \ \ \ \ \ (3)$

$\it (1),\ (2),\ (3) \Longrightarrow f(x)\ continu\breve{a}\ \^{ i}n\ x_{0} =1$

$\it f'(x) = \begin{cases} \it 2x-1,\ x <1\\ \\ \it -2x+1,\ x>1\end{cases}\\ \\ \\ f'(0) = 2\cdot0-1=-1\\ \\ \\ f'(2) = -2\cdot2+1 = -4+1=-3\\ \\ \\ f'(0) +f'(2) = -1-3=-4$