 | ■No23416に返信(恵美さんの記事)
> x≧0のとき、つねにx^3-ax+1≧0が成り立つように実数aの範囲を定めよ。
x>0 で 有理函数 x---->(x^3+1)/x を考察；
In[10]:=
Solve[x^3 - a*x + 1 == 0, a]
Out[10]=
{{a -> -((-1 - x^3)/x)}}
In[11]:=
D[-((-1 - x^3)/x), x]
Out[11]=
3*x + (-1 - x^3)/x^2
In[12]:=
Solve[% == 0, x]
Out[12]=
{{x -> -(-(1/2))^(1/3)}, {x -> 1/2^(1/3)},
{x -> (-1)^(2/3)/2^(1/3)}}
In[13]:=
-((-1 - x^3)/x) /. {x -> 1/2^(1/3)}
Out[13]=
3/2^(2/3)
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