Réponses

2014-01-16T10:43:24+01:00
Bonjour,

V(x)= \dfrac{1}{6}x(8-x)^2\\\\V'(x)= [\dfrac{1}{6}x(8-x)^2]'= \dfrac{1}{6}[x(8-x)^2]'\\\\= \dfrac{1}{6}[x'\times(8-x)^2+((8-x)^2)'\times x]\\\\= \dfrac{1}{6}[1\times(8-x)^2-2(8-x)\times x]\\\\= \dfrac{1}{6}[(8-x)^2-2x(8-x)]\\\\= \dfrac{1}{6}(8-x)[(8-x)-2x]\\\\= \dfrac{1}{6}(8-x)(8-3x)

Signes de la dérivée V'(x) et variation de la fonction V

Racines de V'(x) : 8 - x = 0 ===> x = 8
                                8 - 3x = 0 ===> 3x = 8 ===> x = 8/3

\begin{array}{|c|ccccccc||}x&-\infty&&\dfrac{8}{3}&&8&&+\infty\\ \dfrac{1}{6}(8-x)&&+&+&+&0&-&\\ 8-3x&&+&0&-&-&-&\\ V'x)&&+&0&-&0&+& \\\end{array}

Or  x ∈ [0 ; 8]

\begin{array}{|c|ccccc||}x&0&&\dfrac{8}{3}\approx2,7&&8\\ V'(x)&&+&0&-&0 \\ V(x)&0&\nearrow&\dfrac{1024}{81}\approx12,6&\searrow&0 \end{array}