作业帮已知函数f(x)=e^x+ax^2+bx.设函数f(x)在点(t,f(t))(0
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已知函数f(x)=e^x+ax^2+bx.设函数f(x)在点(t,f(t))(0
已知函数f(x)=e^x+ax^2+bx.设函数f(x)在点(t,f(t))(0
已知函数f(x)=e^x+ax^2+bx.设函数f(x)在点(t,f(t))(0
已知函数f(x)=e^x+ax²+bx.设函数f(x)在点(t,f(t))(0
f'(x)=e^x+2ax+b
所以 l 的斜率 为 e^t+2at+b
同样点（t，f(t))和点Q(0，y)在 l 上，所以l的斜率为 （f(t)-y)/t=e^t/t+at+b-y/t
e^t/t+at+b-y/t=e^t+2at+b
y=e^t-t*e^t-at^2
y'=e^t-t*e^t-e^t-2at=-2at-t*e^t=-t(2a+e^t)...
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f'(x)=e^x+2ax+b
所以 l 的斜率 为 e^t+2at+b
同样点（t，f(t))和点Q(0，y)在 l 上，所以l的斜率为 （f(t)-y)/t=e^t/t+at+b-y/t
e^t/t+at+b-y/t=e^t+2at+b
y=e^t-t*e^t-at^2
y'=e^t-t*e^t-e^t-2at=-2at-t*e^t=-t(2a+e^t)
当a>-1/2是 2a+e^t >0 y' 恒小于0
所以t=0时 y最大
y=e^0-0*e^0-2a0^2=1 成立
当a<-e/2 时 2a+e^t<0 y' 恒大于 0
所以t=1时 y最大
y=e-e-a=-a>e/2>1 不成立
当-e/2
y1=e-e-a=-a
要让y一直小于1则 y1要小于1
-a<1 a>-1
综上 a>-1时 Q纵坐标恒小于1
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The answer is a>=-1/2。
The equation for the tangent line l is
y-f(t)=f'(t)(x-t).
So the y-coordinate y_Q of Q is
f(t)-tf'(t)=e^t+at^2+bt-t(e^t+2at+b)=(1-t)e^t-at^2.
Since y_Q<1, ...
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The answer is a>=-1/2。
The equation for the tangent line l is
y-f(t)=f'(t)(x-t).
So the y-coordinate y_Q of Q is
f(t)-tf'(t)=e^t+at^2+bt-t(e^t+2at+b)=(1-t)e^t-at^2.
Since y_Q<1, we have
a>((1-t)e^t-1)/t^2. (1)
Now consider the function
g(t)=((1-t)e^t-1)/t^2,
which is exactly the right hand side of (1). We compute
g'(t)=-h(t)/t^3,
where
h(t)=e^t*t^2+2*e^t-2*e^t*t-2.
Since h(0)=0 and h'(t)=e^t*t^2>0, we derive g'(t)<0, i.e., g is decreasing in the interval (0,1).
Therefore, the maximal value of g(t) is g(0).
By L'Hôpital's rule, we can deduce
lim_{t \to 0}g(t)=-1/2. (2)
In view of (1) and (2), we find a>=-1/2. This completes the solution.
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