global maximum/minimum,local maximum/minimum ,它们是全局极大值/极小值和局

global maximum/minimum,local maximum/minimum ,它们是全局极大值/极小值和局部极大值/极小值吧?怎样求全局?怎样求局部?
假设这条题:f(x)=(x^1/3)*(2-x)^2/3
为什麼这条题没有全局极大跟极小值?只有局部的?

我表示什么都不懂,不过给你个例子,
题目：
Global maximum and minimum?
Find the global maximum and minimum values of the function
f(x) = e^{-6x} - e^{-4x}
on the interval [1,3].
The global maximum is at x = __.
The global minimum is at x = __.
I'm lost as to how to find this.Can anyone help me through it?
3 years ago
答案1：
Let g = -6x,so dg/dx = -6
Let h = e^g = e^(-6x),so dh/dg = e^g = e^(-6x) = h
By the Chain Rule,dh/dx = dh/dg * dg/dx
dh/dx = e^(-6x) * (-6) = -6e^(-6x)
Let j = -4x,so dj/dx = -4
Let k = e^j = e^(-4x),so dk/dj = e^j = e^(-4x) = k
By the Chain Rule,dk/dx = dk/dj * dj/dx
dk/dx = e^(-4x) * (-4) = -4e^(-4x)
Since f = h - k,we have:
df/dx = dh/dx - dk/dx
df/dx = -6e^(-6x) - (-4e^(-4x))
df/dx = 4e^(-4x) - 6e^(-6x)
To find the maximum/minimum,we set df/dx to zero:
0 = 4e^(-4x) - 6e^(-6x)
6e^(-6x) = 4e^(-4x)
6/4 = e(-4x) / e^(-6x)
3/2 = e^(-4x - (-6x))
3/2 = e^(6x - 4x)
3/2 = e^(2x)
ln(3/2) = 2x
x = ln(3/2) / 2
x 0.2
So there is no maximum or minimum in the interval [1,3]
So within that interval,the maximum/minimum would have to be x=1 or x=3
If x = 1,f(x) -0.0158
If x = 3,f(x) -0.0000061
Of those two,x = 1 is the minimum and x = 3 is the maximum.
3 years ago
答案2：
I will start you out.
Because of the bounds on the interval,you need to compute f(1) and f(3) as they may be either the max of min of the function in this interval.That's for you to do.
Step1.set f'(x) = 0
f'(x) = -6*e^{-6x} +4e^{-4x} = 0
This looks like a mess,but multiple everything by e^(6x) and recall that e^0 = 1
-6 +4*e^(2x) = 0 that's nicer.
e^(2x) = 3/2
take ln of both sides:
2x = ln(3/2) and x = 1/2*ln(3/2)
This may be a global min or max.Compare its value to f(1) and f(3).
To see whether it is a local min or local max,compute the second derivative at this value.
I think that should get you started.