- #51

fresh_42

Mentor

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- 13,636

\begin{align*}Oops, I somehow missed the adjective "real" in the question! I guess I should've realized it was a little too easy.

I still think ##P## can win, so I'll try again. Let ##f(x)=x^3+ax^2+bx+c.## First ##P## chooses ##b=-1##. Note that this ensures that the discriminant of ##f'(x)=3x^2+2ax+b##, which is ##4a^2-12b## must be strictly positive positive,

f'(x)=0&=3x^2+2ax-1 \\&\Longrightarrow 0=x^2+\dfrac{2a}{3}-\dfrac{1}{3}\\&\Longrightarrow x_{1,2}=-\dfrac{a}{3}\pm \sqrt{\dfrac{a^2}{9}+\dfrac{1}{3}}=-\dfrac{a}{3}\pm \dfrac{1}{3}\sqrt{a^2+3}

\end{align*}

which is always a positive discriminant.

so ##f## has two distinct critical points, the smaller of which must be a local maximum, and the larger of which a local minimum (no matter the value of the other coefficients). If ##Q## selects a value for ##b##,

How? ##P## already set ##b=1##.

Hint: It is surprisingly easy if you do not want to enforce a win by the first move!