Question

Fie f:R*->R, f(x)=((x+3)(x-1)^2)/x^2.
a) asimpt oblica spre +infinit.
b) Det punctele de extrem local alr funcției f.
Dacă ajuta am arătat la c) ca f’(x)=((x-1)(x^2+x+6))/x^3

Answer 1

a)

$f(x) = \dfrac{(x+3)(x-1)^2}{x^2} \\ \\ \lim\limits_{x\to \infty}\Big(\dfrac{(x+3)(x-1)^2}{x^2}-mx-n\Big) = 0 \\\\\lim\limits_{x\to \infty}\dfrac{(x+3)(x-1)^2-mx^3-nx^2}{x^2}= 0\\ \\ \lim\limits_{x\to \infty}\dfrac{x^3+x^2-5x+3-mx^3-nx^2}{x^2}= 0\\ \\ \lim\limits_{x\to \infty}\dfrac{x^3(1-m)+x^2(1-n)-5x+3}{x^2}= 0\\ \\ \Rightarrow 1-m = 0\text{ si }1-n = 0\Rightarrow m = 1\text{ si }n = 1 \\ \\ \Rightarrow y = mx+n \Rightarrow\boxed{y = x+1}\to \text{asimptota oblica spre }+\infty$

b)

Egalezi (x-1)(x²+x+6) = 0 => x-1 = 0 sau x²+x+6 = 0 (Δ < 0) =>

=> x = 1 sau x ∈ ∅ => x = 1 punct de extrem local al functiei.