Question

найдите наибольшее значение функции y=x^3-9x^2+24x-1 на отрезке [-1;3]

Answer 1

$\displaystyle f(x)=x^3-9x^2+24x-1;$

$\displaystyle f'(x)=\frac{df(x)}{dx}=\frac{d}{dx}\Big(x^3-9x^2+24x-1\Big)=$
$\displaystyle =\frac{d}{dx}(x^3)-\frac{d}{dx}(9x^2)+\frac{d}{dx}(24x)-\frac{d}{dx}(1)=$
$\displaystyle =3x^2-9\frac{d}{dx}(x^2)+24\frac{dx}{dx}-0=$
$\displaystyle =3x^2-9\cdot 2x+24\cdot 1=3x^2-18x+24;$

$\displaystyle f'(x)=0;$
$\displaystyle 0=3x^2-18x+24;$
$\displaystyle x=\frac{-(-18)\pm\sqrt{(-18)^2-4\cdot 3\cdot 24}}{2\cdot 3}=\frac{18\pm\sqrt{36}}{6}=3\pm 1;$
$\displaystyle x=2\lor x=4;$

$\displaystyle f(-1)=(-1)^3-9(-1)^2+24(-1)-1=-35;$
$\displaystyle f(2)=2^3-9\cdot 2^2+24\cdot 2-1=19;$
$\displaystyle f(3)=3^3-9\cdot 3^2+24\cdot 3-1=17;$

$\displaystyle \max_{x\in{[-1;3]}}f(x)=f(2)=\boxed{19}\phantom{.}.$