`y=(4)/(1+x^2)`
TXĐ: `D=RR`
`y'=(-4.2x)/(1+x^2)^2`
`=(-8x)/(1+x^2)^2`
`y'=0<=>-8x=0`
`<=>x=0`
BBT:
\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}&\text{0}&&+\infty\\\hline \text{$y'$}&\text{}+&\text{}&0&-&\\\hline \text{$y$}&\text{}&\text{}&4\\&&\nearrow&&\searrow\\&-\infty&&&&-\infty\\\hline \end{array}
Vậy `max\text( )y=y(0)=4`
`\text( )`
`y=4x^3-3x^4`
TXĐ: `D=RR`
`y'=12x^2-12x^3`
`y'=0<=>12x^2-12x^3=0`
`<=>12x^2.(1-x)=0`
`<=>`\(\left[ \begin{array}{l}x=0\text{( nghiệm kép)}\\x=1\end{array} \right.\)
BBT:
\begin{array}{|c|cc|}\hline \text{$x$}&\text{$-\infty$}&\text{}0&&\text{1}&+\infty\\\hline \text{$y'$}&\text{}+&\text{}0&+&0&-\\\hline \text{$y$}&\text{}&&\text{}&1\\&&&\nearrow&&\searrow\\&-\infty&&&&-\infty\\\hline \end{array}
Vậy `max\text( )y=y(1)=1`
