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y=e^x+4e^-2xfind the coordinate and nature of the stationary points.
(ans: (ln 2,3) minimum point)
y=x^2 ln x
find all the points of inflexion.
(ans: e^-2/3 , -3e^-3/2) |
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发表于 14-8-2010 12:36 PM
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显示全部楼层
y=e^x+4e^-2x
dy/dx = e^x -8e^(-2x)
dy/dx=0
e^x -8e^(-2x)=0
e^x=8e^(-2x)
x ln e = ln 8 x^(-2x)
= -2x ln e + ln 8
3x = ln 8
x= (1/3)(ln8)
= ln 8^(1/3)
= ln 2
when x = ln 2 , y =e^ln 2 +4e^-2ln 2
= 2 +4(1/4)
=3
dy/dx = e^x -8e^(-2x)
d^2y/dx^2 = e^x + 16e^(-2x)
when ,x=ln2 ,d^2y/dx^2 >0
therefore (ln 2,3) is a minimum pnt
2.y=x^2 ln x
dy/dx = x^2(1/x) + (2x) ln x
= x +(2x) ln x
d^2y/dx^2 = 1 + 2x(1/x) + 2ln x
d^2y/dx^2 = 0
1 + 2x(1/x) + 2ln x=0
lnx = -3/2
x = e^(-3/2)
y=x^2 ln x
y= (e^(-3/2))^2 ln e^(-3/2)
= e^-3 (-3/2)
o.0 答案有一点出入 |
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