# User:Cristianmejia

2):${\displaystyle \ f(x)=x^{4}-27x^{2}}$

${\displaystyle \ f'(x)=4x^{3}-54x}$
${\displaystyle \ 0=4x^{3}-54x}$
${\displaystyle \ 0=x(4x^{2}-54)}$
${\displaystyle \ x_{1}=0}$
${\displaystyle \ x_{2}=4x^{2}-54}$
${\displaystyle \ 4x^{2}=54}$
${\displaystyle \ x^{2}={\frac {54}{4}}}$
${\displaystyle \ x_{2}=+{\sqrt[{2}]{(}}54/4)=3.67}$
${\displaystyle \ x_{2}=-{\sqrt[{2}]{(}}54/4)=-3.67}$
${\displaystyle \ f''(x)=12x^{2}-54}$
${\displaystyle \ f''(0)=12(0)^{2}-54}$
${\displaystyle \ f''(0)=-54}$ Máximo
${\displaystyle \ f''(-3.67)=12(-3.67)^{2}-54=107.6}$ Minimo
${\displaystyle \ f''(+3.67)=12(+3.67)^{2}-54=107.6}$ Minimo

3):${\displaystyle \ f(x)=x^{3}-4x^{2}+4x-1}$

${\displaystyle \ f'(x)=3x^{2}-8x^{+}4}$
${\displaystyle \ 0=3x^{2}-8x^{+}4}$
${\displaystyle \ 0=(x-2)(3x-2)}$
${\displaystyle \ x_{1}=2}$
${\displaystyle \ x_{2}=3x-3}$
${\displaystyle \ x_{2}=2/3}$

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

${\displaystyle \lim _{x\to -3}{\frac {x^{2}-x-12}{x+3}}}$

RESPUESTA:
evaluando:
${\displaystyle \lim _{x\to -3}{\frac {x^{2}-x-12}{x+3}}}$

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

${\displaystyle \lim _{x\to -3}{\frac {x^{2}-x-12}{x+3}}}$

=${\displaystyle \lim _{x\to -3}{\frac {(x+3)(x-4)}{x+3}}}$ =${\displaystyle \lim _{x\to -3}{x-4}=-7}$

${\displaystyle \lim _{x\to 0}{\frac {{\sqrt {(}}2-t)-{\sqrt {2}}}{t}}}$

RESPUESTA:
evaluando:

${\displaystyle \lim _{x\to 0}{\frac {{\sqrt {(}}2-t)-{\sqrt {2}}}{t}}}$ =${\displaystyle {\frac {{\sqrt {(}}2-0)-{\sqrt {2}}}{0}}}$ =${\displaystyle {\frac {{\sqrt {2}}-{\sqrt {2}}}{0}}={\frac {0}{0}}}$

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

${\displaystyle \lim _{x\to 0}{\frac {x}{{\sqrt {(}}1+3x)-1}}}$

RESPUESTA:
evaluando:

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

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

${\displaystyle \lim _{x\to 1}{\frac {{\sqrt[{3}]{x}}-1}{{\sqrt[{2}]{x}}-1}}}$

RESPUESTA:
evaluando:

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

${\displaystyle \lim _{x\to 1}{\frac {{\sqrt[{3}]{x}}-1}{{\sqrt[{2}]{x}}-1}}}$