设f(x)=2^x-x^2-1;
f‘(x)=ln2*2^x-2x;
f''(x)=ln2*ln2*2^x-2;单调,只有一个零点.
故f'(x)至多有两个零点.(roll定理,每两个零点间都有一个导数的零点)
所以f(x)至多三个零点.(理由同上)
当x趋于负无穷时f趋于负无穷.
f(0)=0,f(1)=0,f(2)=-1
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