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求函数极值点和极值(1)y=4-3x-x2 (2)f(x)=-4x3+3x2+6x+2 (3)f(x)=x2e(-x) (4)f(x)=2x3+3x2-12x-1
要先求定义域!
人气:277 ℃ 时间:2020-05-08 11:27:21
解答
所有函数中只包含代数多项式和e^(-x),没有除法,定义域全部为R
1)f'(x) = -3-2x = 0,x = -3/2 f(-3/2) = 4 + 9/2 - 9/4 = 25/4
2)f'(x) = -12x²+6x+6=0,x1=-1/2,x2=1,
f''(x) = -24x+6,f''(1)≠0,f''(-1/2)≠0,即x = -1/2,1均为极值点
f(-1/2) = 4/8+3/4-3+2=1/4,f(1) = -4+3+6+2=7
3)f'(x) = (2x-x²)e^(-x) = 0,x1=0,x2=2,
f''(x) = (2-4x+x²)e^(-x),f(0)≠0,f(2)≠0,
即x = 2,0均为极值点f(0) = 0,f(2) =4/e²
4)f'(x) = 6x²+6x-12 = 0,x1=1,x2=-2,
f''(x) = 12x-6,f''(1)≠0,f''(-2)≠0
即x = -2,1均为极值点
f(1) = 2+3-12-1 = -8,f(-2) = -16+12+24-1=19e^(-x),中^表示什么意思乘方,matlab中乘方符号,在没有公式编辑器条件下大家都这么写你的乘方都没写
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