微积分问题,急~! 要过程 30分 好的再加20!速度啊
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.
赞 peace_king 幼苗
共回答了12个问题采纳率：100% 举报
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!
1年前
7
可能相似的问题
-
1年前8个回答
-
1年前1个回答
-
1年前2个回答
-
1年前3个回答
-
1年前2个回答
-
1年前2个回答
-
1年前3个回答
你能帮帮他们吗