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

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