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

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 functionf(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 = -6Let h = e^g = e^(-6x),so dh/dg = e^g = e^(-6x) = hBy the Chain Rule,dh/dx = dh/dg * dg/dxdh/dx = e^(-6x) * (-6) = -6e^(-6x)Let j = -4x,so dj/dx = -4Let k = e^j = e^(-4x),so dk/dj = e^j = e^(-4x) = kBy the Chain Rule,dk/dx = dk/dj * dj/dxdk/dx = e^(-4x) * (-4) = -4e^(-4x)Since f = h - k,we have:df/dx = dh/dx - dk/dxdf/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) = 2xx = ln(3/2) / 2x 0。2So 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=3If x = 1,f(x) -0。0158If x = 3,f(x) -0。0000061Of 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) = 0f'(x) = -6*e^{-6x} +4e^{-4x} = 0This 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) = 3/2take 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。