回答:
说明:
确定的衍生物
取分子然后等于零
简化
考虑常用术语
x的值是:
使用
使用
什么是f(x)= 4x ^ 2-2x + x /(x-1/4)的局部极值?
F_(分钟)= F(1/4 + 2 ^( - 5/3))=(2 ^(2/3)+ 3 + 2 ^(5/3))/ 4。观察到,f(x)= 4x ^ 2-2x + x /(x-1/4); x在RR- {1/4}。 = 4X ^ 2-2x + 1 / 4-1 / 4 + {(X-1/4)+1/4} /(X-1/4); xne1 / 4 =(2x-1/2)^ 2-1 / 4 + {(x-1/4)/(x-1/4)+(1/4)/(x-1/4)}; xne1 / 4 = 4(x-1/4)^ 2-1 / 4 + {1+(1/4)/(x-1/4)}; xne1 / 4 :. F(X)= 4(X-1/4)^ 2 + 3/4 +(1/4)/(X-1/4); xne1 / 4。现在,对于局部极值,f'(x)= 0,和f''(x)>或<0,“根据”f_(min)或f_(max),“resp。” f'(x)= 0 rArr 4 {2(x-1/4)} + 0 + 1/4 {( - 1)/(x-1/4)^ 2} = 0 ...(ast)rArr 8(x-1/4)= 1 / {4(x-1/4)^ 2},或(x-1/4)^ 3 = 1/32 = 2 ^ -5。 rArr x = 1/4 + 2 ^( - 5/3)此外,(ast)rArr f''(x)= 8-1 / 4 {-2(x-1/4)^ - 3},“所以那个,“f
什么是f(x)= x ^ 2 /(x ^ 2-3x-5)的局部极值?
MAX(0; 0)和MIN(-10 / 3,20 / 29)我们计算f'(x)= - x(3x + 10)/(x ^ 2-3x-5)^ 2 f''(x )= 2(3x ^ 2 + 15x ^ 2 + 25)/(x ^ 2-3x-5)^ 3所以f'(x)= 0如果x = 0或x = -10 / 3我们还有f' '(0)= - 2/5 <0且f''( - 10/3)= 162/4205> 0
什么是f(x)=((x-2)(x-4)^ 3)/(x ^ 2-2)的局部极值?
X = -5 f(x)= [(x-2)(x-4)^ 3] /(x ^ 2-2)x ^ 2-2 =(x + 2)(x-2)所以函数将成为:f(x)= [(x-4)^ 3] /(x + 2)现在f'(x)= d / dx [(x-4)^ 3] /(x + 2)f' (x)= [3(x + 2)(x-4)^ 2-(x-4)^ 3] /(x + 2)^ 2对于局部极值点f'(x)= 0所以[3( x + 2)(x-4)^ 2-(x-4)^ 3] /(x + 2)^ 2 = 0 [3(x + 2)(x-4)^ 2-(x-4) ^ 3] = 0 3(x + 2)(x-4)^ 2 =(x-4)^ 3 3x + 6 = x-4 2x = -10 x = -5