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微积分问题,急~!要过程30分好的再加20!速度啊1.findtheabsolutemaximumandabsoluteminimumofthefunctiong(x)=x/5+5/xovertheclosedinterval[1/100,100]2.findtwopositivenumberssuchthatthesumofthefirstandthre
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微积分问题,急~! 要过程 30分 好的再加20!速度啊
1. find the absolute maximum and absolute minimum of the function g(x)=x/5 + 5/x over the closed interval[ 1/100, 100]
2. find two positive numbers such that the sum of the first and three times the second is 240 and the product is a maximum.
3. find the equation of the tangent line T(x) to the graph of f(x)=√x(根号x) at the point(16, 4). use T(x)to approximate the value of √ 17(根号17). simplify your answer by hand(yes, do the long division)- you should end up with a real number with 3 decimal places.
1. find the absolute maximum and absolute minimum of the function g(x)=x/5 + 5/x over the closed interval[ 1/100, 100]
2. find two positive numbers such that the sum of the first and three times the second is 240 and the product is a maximum.
3. find the equation of the tangent line T(x) to the graph of f(x)=√x(根号x) at the point(16, 4). use T(x)to approximate the value of √ 17(根号17). simplify your answer by hand(yes, do the long division)- you should end up with a real number with 3 decimal places.
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答案和解析
1.函数g(x)的最大值在x=100时,为20.05;最小值在x=5时,为2.
求g(x)的导数g'(x)=(x^2-25)/[5*(x^2)].于是可知,在x5时,g(x)单调上升.
2.设两个正数x,y.则依题意有方程:x+3y=240,并且方程h(x)=3xy取得最大值.
x=240-3y,代入h(x),求h(x)的导数h'(x)=720-18y,知当y=40时符合h(x)取得最大值的条件,此时x=120.即两个数分别为120,40.
3.f(x)=√x,则其导函数为:f'(x)=1/(2*√x),在点(16,4)处,f'(x)=1/8=0.125.于是切线方程为:T(x)=0.125(x-16)+4=0.125x+2.最总用T(x)来近似表示√ 17=0.125*17+2=4.125,与真实值4.1231056相差不大.
要给分哦,寝室11电都停电了,我是用笔记本电池帮你做的这些题目!
求g(x)的导数g'(x)=(x^2-25)/[5*(x^2)].于是可知,在x5时,g(x)单调上升.
2.设两个正数x,y.则依题意有方程:x+3y=240,并且方程h(x)=3xy取得最大值.
x=240-3y,代入h(x),求h(x)的导数h'(x)=720-18y,知当y=40时符合h(x)取得最大值的条件,此时x=120.即两个数分别为120,40.
3.f(x)=√x,则其导函数为:f'(x)=1/(2*√x),在点(16,4)处,f'(x)=1/8=0.125.于是切线方程为:T(x)=0.125(x-16)+4=0.125x+2.最总用T(x)来近似表示√ 17=0.125*17+2=4.125,与真实值4.1231056相差不大.
要给分哦,寝室11电都停电了,我是用笔记本电池帮你做的这些题目!
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