求 f ( x ) f(x) f(x)在 [ a , b ] [a,b] [a,b]上最大值和最小值的方法:
求 f ( x ) = ( x − 5 ) x 2 3 f(x)=(x-5)\sqrt[3]{x^2} f(x)=(x−5)3x2 在 [ − 2 , 3 ] [-2,3] [−2,3]上的最值
解: f ′ ( x ) = x 2 3 + 2 3 ( x − 5 ) 1 x 3 = 5 ( x − 2 ) 3 x 3 f'(x)=\sqrt[3]{x^2}+\dfrac 23(x-5)\dfrac{1}{\sqrt[3]{x}}=\dfrac{5(x-2)}{3\sqrt[3]{x}} f′(x)=3x2 +32(x−5)3x 1=33x 5(x−2)
可能极值点: x 1 = 0 , x 2 = 2 x_1=0,x_2=2 x1=0,x2=2
f ( 0 ) = 0 , f ( 2 ) = − 3 4 3 , f ( − 2 ) = − 7 4 3 , f ( 3 ) = − 2 9 3 f(0)=0,f(2)=-3\sqrt[3]4,f(-2)=-7\sqrt[3]4,f(3)=-2\sqrt[3]9 f(0)=0,f(2)=−334 ,f(−2)=−734 ,f(3)=−239
所以最大值为 0 0 0,最小值为 − 7 4 3 -7\sqrt[3]4 −734
函数 f ( x ) = 1 3 x 3 − 2 x 2 + 5 f(x)=\dfrac 13x^3-2x^2+5 f(x)=31x3−2x2+5在 [ − 2 , 2 ] [-2,2] [−2,2]的最大值为 ‾ \underline{\qquad}
解: f ′ ( x ) = x 2 − 4 x = x ( x − 4 ) f'(x)=x^2-4x=x(x-4) f′(x)=x2−4x=x(x−4)
可能的极值点: x 1 = 0 , x 2 = 4 x_1=0,x_2=4 x1=0,x2=4
f ( 0 ) = 5 , f ( 4 ) = − 17 3 , f ( − 2 ) = − 17 3 , f ( 2 ) = − 1 3 f(0)=5,f(4)=-\dfrac{17}{3},f(-2)=-\dfrac{17}{3},f(2)=-\dfrac 13 f(0)=5,f(4)=−317,f(−2)=−317,f(2)=−31
所以最大值为 5 5 5