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

[latex]\displaystyle f(x)=x^3-9x^2+24x-1;[/latex] [latex]\displaystyle f'(x)=\frac{df(x)}{dx}=\frac{d}{dx}\Big(x^3-9x^2+24x-1\Big)=[/latex] [latex]\displaystyle =\frac{d}{dx}(x^3)-\frac{d}{dx}(9x^2)+\frac{d}{dx}(24x)-\frac{d}{dx}(1)=[/latex] [latex]\displaystyle =3x^2-9\frac{d}{dx}(x^2)+24\frac{dx}{dx}-0=[/latex] [latex]\displaystyle =3x^2-9\cdot 2x+24\cdot 1=3x^2-18x+24;[/latex] [latex]\displaystyle f'(x)=0;[/latex] [latex]\displaystyle 0=3x^2-18x+24;[/latex] [latex]\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;[/latex] [latex]\displaystyle x=2\lor x=4;[/latex] [latex]\displaystyle f(-1)=(-1)^3-9(-1)^2+24(-1)-1=-35;[/latex] [latex]\displaystyle f(2)=2^3-9\cdot 2^2+24\cdot 2-1=19;[/latex] [latex]\displaystyle f(3)=3^3-9\cdot 3^2+24\cdot 3-1=17;[/latex] [latex]\displaystyle \max_{x\in{[-1;3]}}f(x)=f(2)=\boxed{19}\phantom{.}.[/latex]