Question

Va rog putin ajutor .....

Answer 1

Răspuns:

m=10

Explicație pas cu pas:

f(x)=2x²+4x, m=f'(x_{0})=6*x_{0}+4. f( x_{0})=2*x_{0} ^{3}+4* x_{0}.

Ecuatia tangentei :  y=f(x_{0})+f'(x_{0})*(x-x_{0}), de unde

y=f'(x_{0})*x+f(x_{0})-f'(x_{0})*x_{0}, deci f(x_{0})-f'(x_{0})*x_{0}=4

Dupa inlocuire avem: 2*x_{0} ^{3}+4* x_{0}-(6*x_{0}+4)x_{0}=4

de unde obtinem -4x_{0}^{3}=4. Deci x_{0}=-1 si deci m=6+4=10

Answer 2

$\it f(x) =2x^3+4x \Rightarrow f'(x) =6x^2+4\\ \\ Fie\ T(x_0,\ y_0) \ punctul\ de\ tangen\c{\it t}\breve{a} \Rightarrow m=f'(x_0)=6x_0^2+4\ \ \ \ (1)\\ \\ y=mx+4\Rightarrow y_0=mx_0+4\stackrel{(1)}{\Longrightarrow}\ y_0=(6x_0^2+4)x_0+4\ \Rightarrow \\ \\ \Rightarrow y_0=6x_0^3+4x_0+4\ \ \ \ (2)$

$\it T(x_0,\ y_0)\in Gf \Rightarrow y_0=2x_0^3+4x_0\ \ \ \ (3)\\ \\ (2),\ (3) \Rightarrow 6x_0^3+4x_0+4=2x_0^3+4x_0 \Rightarrow 4x_0^3 +4=0|_{:4} \Rightarrow x_0^3+1=0 \Rightarrow \\ \\ \Rightarrow x_0^3=-1 \stackrel{x\in \mathbb{R}}{\Longrightarrow}\ x_0=-1\ \ \ \ (4)\\ \\ (1),\ (2) \Rightarrow m=6\cdot(-1)^2+4=6+4=10$