1: \(y=x^3-3x^2-9x+2\)
=>y'=\(3x^2-3\cdot2x-9=3x^2-6x-9=3\left(x^2-2x-3\right)=3\left(x-3\right)\left(x+1\right)\)
Đặt y'=0
=>(x-3)(x+1)=0
=>x=3(loại) hoặc x=-1(nhận)
\(f\left(-2\right)=\left(-2\right)^3-3\cdot\left(-2\right)^2-9\cdot\left(-2\right)+2\)
=-8-12+18+2
=-20+20
=0
\(f\left(2\right)=2^3-3\cdot2^2-9\cdot2+2\)
=8-18-18+2
=10-36
=-26
\(f\left(-1\right)=\left(-1\right)^3-3\cdot\left(-1\right)^2-9\cdot\left(-1\right)+2\)
=-1-3+9+2
=-4+11
=7
Vậy: \(f\left(x\right)_{\max\left\lbrack-2;2\right\rbrack}\) =7 khi x=-1
\(f\left(x\right)_{\min\left\lbrack-2;2\right\rbrack}\) =-26 khi x=2
Bài 2: \(f\left(x\right)=x^4-2x^2+1\)
=>f'(x)=\(4x^3-2\cdot2x=4x\left(x^2-1\right)=4x\left(x-1\right)\left(x+1\right)\)
Đặt f'(x)=0
=>x(x-1)(x+1)=0
=>x=0(nhận) hoặc x=1(nhận) hoặc x=-1(loại)
\(f\left(0\right)=0^4-2\cdot0^2+1=1\)
\(f\left(1\right)=1^4-2\cdot1^2+1=0\)
\(f\left(2\right)=2^4-2\cdot2^2+1=16-8+1=9\)
Vậy: \(f\left(x\right)_{\max\left\lbrack0;2\right\rbrack}\) =9 khi x=2
3: \(f\left(x\right)=x^3-8x^2+16x-9\)
=>f'(x)=\(3x^2-8\cdot2x+16=3x^2-16x+16\) =3x^2-4x-12x+16
=x(3x-4)-4(3x-4)
=(3x-4)(x-4)
Đặt f'(x)=0
=>(3x-4)(x-4)=0
=>x=4/3(nhận) hoặc x=4/3(loại)
\(f\left(1\right)=1^3-8\cdot1^2+16-9=1-8+16-9=0\)
\(f\left(\frac43\right)=\left(\frac43\right)^3-8\cdot\left(\frac43\right)^2+16\cdot\frac43-9\)
\(=\frac{64}{27}-\frac{128}{9}+\frac{64}{3}-9=\frac{13}{27}\)
\(f\left(3\right)=3^3-8\cdot3^2+16\cdot3-9=-6\)
Vậy: \(f\left(x\right)_{\max\left\lbrack1;3\right\rbrack}\) =13/27 khi x=4/3
