Question

Se considera functia f:R -> R,f(x)= -2x+1 supra 4 ,x <0
1 supra x la 2 + 4 ,x >=0
sa se studieze continuitatea functiei f in punctul x0 = 0
sa se scrie ecuatia tangentei la graficul functiei f in punctul de abscisa 1
sa se demonstreze ca functia f este descrescatoare pe intervalul (0, infinit)

Answer 1

$\displaystyle f(x)=\left \{ {{\displaystyle\frac{-2x+1}4, \:x<0} \atop {\displaystyle\frac1{x^2+4},\: x\geq0}} \right. \\a)\:\text{Studiem continuitatea func\c tiei }f \text{ \^in punctul }x_0=0\text{ cu limite laterale:}\\\lim\limits_{x\to0,\:x<0}f(x)=\lim\limits_{x\to0,\:x<0}\displaystyle\frac{-2x+1}{4}=\frac14\\\lim\limits_{x\to0,\:x>0}f(x)=\lim\limits_{x\to0,\:x>0}\frac{1}{x^2+4}=\frac14=f(0)\\\Rightarrow f\text{ continu\u a \^in }x_0=0$

$b)\:\displaystyle f'(x)=\left \{ {{\displaystyle\left(\frac{-2x+1}4\right)', \:x<0} \atop {\displaystyle\left(\frac1{x^2+4}\right)',\: x>0}} \right.\\\\\displaystyle f'(x)=\left \{ {{\displaystyle-\frac12, \:x<0} \atop {\displaystyle\frac{-2x}{(x^2+4)^2},\: x>0}} \right.\\\\\\t:y-f(1)=f'(1)(x-1)\\\\\displaystyle f(1)=\frac{1}{1+4}=\frac15\\f'(1)=\frac{-2}{(1+4)^2}=\frac{-2}{25}\\\\t:y-\frac15=\frac{-2}{25}(x-1)\\\\t:y=\frac{-2}{25}x+\frac{7}{25}$

$c)\: \displaystyle f'(x)=\frac{-2x}{(x^2+4)^2}<0,\forall x \in(0,\infty)\Rightarrow f\text{ descresc\u atoare pe }(0, \infty)$