$y'=2(x+3)^2+2x\times 2(x+3)$
$y'=2(x^2+6x+9)+4x^2+12x$
$y'=2x^2+12x+18+4x^2+12x$
$y'=6x^2+24x+18$
$\therefore 6x^2+24x+18=14$$6x^2+24x+4=0$
$3x^2+12x+2=0$
$x=\displaystyle { -12\pm \sqrt{12^2-4(3)(2)} \over 6}$
$x=\displaystyle { -12\pm \sqrt{120} \over 6}$
$x=\displaystyle { -12\pm 2\sqrt{30} \over 6}$
$x=\displaystyle { -6\pm \sqrt{30} \over 3}$
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