User:Cristianmejia

2):$$\ f(x) = x^4-27x^2$$
 * $$\ f'(x) = 4x^3-54x$$
 * $$\ 0 = 4x^3-54x$$
 * $$\ 0 = x(4x^2-54)$$
 * $$\ x_1 = 0$$
 * $$\ x_2 = 4x^2-54$$
 * $$\ 4x^2 = 54$$
 * $$\ x^2 = \frac{54}{4}$$
 * $$\ x_2 = +\sqrt[2](54/4)=3.67$$
 * $$\ x_2 = -\sqrt[2](54/4)=-3.67$$


 * $$\ f''(x) = 12x^2-54$$
 * $$\ f''(0) = 12(0)^2-54$$
 * $$\ f''(0) = -54$$ Máximo


 * $$\ f''(-3.67) = 12(-3.67)^2-54=107.6 $$ Minimo
 * $$\ f''(+3.67) = 12(+3.67)^2-54=107.6$$ Minimo

cristianmejia

3):$$\ f(x) = x^3-4x^2+4x-1$$
 * $$\ f'(x) = 3x^2-8x^+4$$
 * $$\ 0 = 3x^2-8x^+4$$
 * $$\ 0 = (x-2)(3x-2)$$


 * $$\ x_1 = 2$$
 * $$\ x_2 = 3x-3$$
 * $$\ x_2 = 2/3 $$


 * $$\ f''(x) = 6x-8$$
 * $$\ f''(2) = 6(2)-8= 4$$ Minimo
 * $$\ f''(2/3) = 6(2/3)-8= 1$$ Minimo

cristianmejia

$$\lim_{x\to -3}\frac {x^2-x-12}{x+3}$$


 * RESPUESTA:
 * evaluando:
 * $$\lim_{x\to -3}\frac {x^2-x-12}{x+3}$$

=$$ \frac {(-3)^2-(-3)-12}{(-3)+3}$$ =$$ \frac {9+3-12}{0}$$ =$$ \frac {0}{0}$$


 * $$\lim_{x\to -3}\frac {x^2-x-12}{x+3}$$

=$$\lim_{x\to -3}\frac {(x+3)(x-4)}{x+3}$$ =$$\lim_{x\to -3}{x-4}=-7$$

cristianmejia

$$\lim_{x\to 0}\frac {\sqrt (2-t)-\sqrt 2}{t}$$


 * RESPUESTA:
 * evaluando:

$$\lim_{x\to 0}\frac {\sqrt (2-t)-\sqrt 2}{t}$$ =$$\frac {\sqrt (2-0)-\sqrt 2}{0}$$ =$$\frac {\sqrt 2-\sqrt 2}{0}=\frac {0}{0}$$

$$\lim_{x\to 0}\frac {\sqrt (2-t)-\sqrt 2}{t}$$ =$$\lim_{x\to 0}\frac {(\sqrt (2-t)-\sqrt 2)(\sqrt (2-t)+\sqrt 2}{t(\sqrt (2-t)+\sqrt 2}$$ =$$\lim_{x\to 0}\frac {(\sqrt (2-t))^2-(\sqrt 2)^2}{t(\sqrt (2-t)+\sqrt 2)}$$ =$$\lim_{x\to 0}\frac {(2-t)-2)}{t(\sqrt (2-t)+\sqrt 2)}$$ =$$\lim_{x\to 0}\frac {-t}{t(\sqrt (2-t)+\sqrt 2)}$$ =$$\lim_{x\to 0}\frac {-1}{\sqrt (2-t)+\sqrt 2}=\frac {-1}{\sqrt (2-0)+\sqrt 2}$$ =$$\frac {-1}{\sqrt (2)+\sqrt 2}$$ =$$\frac {-1}{2\sqrt 2}$$

cristianmejia

$$\lim_{x\to 0}\frac {x}{\sqrt (1+3x)-1}$$
 * RESPUESTA:
 * evaluando:

$$\lim_{x\to 0}\frac {x}{\sqrt (1+3x)-1}$$ =$$\frac {0}{\sqrt (1+3(0))-1}$$ =$$\frac {0}{\sqrt 1-1}$$ =$$\frac {0}{0}$$

$$\lim_{x\to 0}\frac {x}{\sqrt (1+3x)-1}$$ =$$\lim_{x\to 0}\frac {x(\sqrt (1+3x)+1)}{(\sqrt (1+3x)-1)(\sqrt (1+3x)+1)}$$ =$$\lim_{x\to 0}\frac {x(\sqrt (1+3x)+1)}{(\sqrt (1+3x))^2(1)^2}$$ =$$\lim_{x\to 0}\frac {x(\sqrt (1+3x)+1)}{(1+3x)-(1)}$$ =$$\lim_{x\to 0}\frac {x(\sqrt (1+3x)+1)}{3x)}$$ =$$\lim_{x\to 0}\frac {(\sqrt (1+3x)+1)}{3}=\frac {(\sqrt (1+3(0))+1)}{3}$$ =$$\frac {(\sqrt 1+1)}{3}=\frac {2}{3}$$

cristianmejia

$$\lim_{x\to 1}\frac {\sqrt[3]x-1}{\sqrt[2]x-1}$$
 * RESPUESTA:
 * evaluando:

$$\lim_{x\to 1}\frac {\sqrt[3]x-1}{\sqrt[2]x-1}$$= $$\frac {\sqrt[3]1-1}{\sqrt[2]1-1}$$= $$\frac {1-1}{1-1}$$= $$\frac {0}{0}$$ Indeterminado

$$\lim_{x\to 1}\frac {\sqrt[3]x-1}{\sqrt[2]x-1}$$