Respostas

2014-05-03T13:26:23-03:00
f(x)=x^2-5x

x^2-5x=0\\x(x-5)=0\\x'=0\\
\\x''=5


Deriva e iguala a zero  para determinar o ponto crítico

y=x^2-5x\\y'=2x-5\\
\\
\\0=2x-5\\
\\2x=5\\
\\x= \frac{5}{2}

f( \frac{5}{2} )=( \frac{5}{2} )^2- \frac{5.5}{2} \\
\\f( \frac{5}{2} )= \frac{25}{4} - \frac{25}{2} \\
\\f( \frac{5}{2} )= -\frac{25}{4}

\boxed{V( \frac{5}{2}~:- \frac{25}{4}  )}
1 3 1