作业帮在区间[0,1]上任取两个实数a,b,则函数f(x)=1/2*x^3+ax-b在区间[-1,1]上有且仅有一个零点的概率是多少怎样用图像表示,说的具体点
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在区间[0,1]上任取两个实数a,b,则函数f(x)=1/2*x^3+ax-b在区间[-1,1]上有且仅有一个零点的概率是多少怎样用图像表示,说的具体点
在区间[0,1]上任取两个实数a,b,则函数f(x)=1/2*x^3+ax-b在区间[-1,1]上有且仅有一个零点的概率是多少
怎样用图像表示,说的具体点
在区间[0,1]上任取两个实数a,b,则函数f(x)=1/2*x^3+ax-b在区间[-1,1]上有且仅有一个零点的概率是多少怎样用图像表示,说的具体点
firstly, you want to know what the curve looks like:
f'(x) = 3/2*x^2+a
since a>=0, f'(x) >= 0. And only when x=0 and a=0, f'(x)=0.
So for any a, b in [0.1], function f(x) monotonically increasing. In other words, f(x) has unique zero within [-1,1] if and only if f(-1) <= 0 and f(1) >=0.
since f(-1)=-1/2-a-b<0 is always true when a,b in [0,1], we conclude that:
f(x) has unique zero within [-1,1] <==> f(1) = 1/2+a-b >=0
Now we can play in a figure. When x-axis is a and y-axis is b, we will plot dot on the unit square. Let's plot red dot when (a,b) satisfies that 1/2+a-b >=0 and plot blue dot otherwise. We got a figure like the one attached.
To calculate the probability of 1/2+a-b>=0 is the same as to measure the area of red dots, which is 7/8.
即是求a-b>0的概率。注意在[-1,1]上f'(x)>=0，所以f(x)在[-1,1]上单增，而由于f(0)=-b<=0，所以只要f(1)=a-b>0即对应只有一个解。