zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的函数 >

extreme f(x)=-2x^3+39x^2-240,4<= x<= 9

解答

极值点

f (x) = - 2 x^{3} + 39 x^{2} - 240, 4 \leq x \leq 9

解答

极小值 (4, 256), 极大值 (9, 1461)

求解步骤

找到临界点:

x = 4, x = 9

- 2 x^{3} + 39 x^{2} - 240, 4 \leq x \leq 9

的定义域

: 4 \leq x \leq 9

Find intervals:

递增

: 4 < x < 9

将极值点

x = 4

代入

- 2 x^{3} + 39 x^{2} - 240 \Rightarrow y = 256

极小值 (4, 256)

将极值点

x = 9

代入

- 2 x^{3} + 39 x^{2} - 240 \Rightarrow y = 1461

极大值 (9, 1461)

极小值 (4, 256), 极大值 (9, 1461)

作图

流行的例子

extreme f(x)=8(x^2)/((x-6)),-3<= x<= 2

e x t re m e f (x) = 8 \frac{x ^{2}}{( x - 6 )}, - 3 \leq x \leq 2

extreme f(x)=-0.05x^2+300x+2000

e x t re m e f (x) = - 0.05 x^{2} + 300 x + 2000

extreme f(x,y)=e^{x^2}+y^2+1

e x t re m e f (x, y) = e^{x^{2}} + y^{2} + 1

extreme f(x)=-x^{(3)}+12x^{(2)}+45x-52

e x t re m e f (x) = - x^{(3)} + 12 x^{(2)} + 45 x - 52

extreme f(x)=2cos(x)-1

e x t re m e f (x) = 2 cos (x) - 1