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Let $f(x) = a_1 + b_1 x + c_1 x^2 + d_1 \sqrt{a_2 + b_2 x + c_2 x^2}$ be a function whose domain is $[0,1]$. It is known that $d_1 < 0$ and $\forall x \in [0,1]$, $a_2 + b_2 x + c_2 x^2 > 0$.

What is the maximum number of interior local optima of this function?

Rittwik Chatterjee
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  • Any clue from your side? – Cesareo Dec 01 '22 at 09:45
  • $a_1+b_1 x+c_1 x^2$ has at most one optima. And I think $\sqrt{a_2+b_2 x+c_2 x^2}$ has another. So my guess is that $f(x)$ has at most two local optima. But I don't have any proof. – Rittwik Chatterjee Dec 02 '22 at 10:48
  • Did you try writing the first-order optimality condition $f'(x)=0$? – Kas Dec 02 '22 at 15:58
  • Yes. It is $(b_1 + c_1 x)+ \frac{d_1(b_2 + 2c_2 x)}{2 \sqrt{a_2 + b_2 x + c_2 x^2}} = 0$. This means $2(b_1 + c_1 x)\sqrt{a_2 + b_2 x + c_2 x^2} = - d_1(b_2 +2 c_2 x)$. If I square both sides I get a polynomial of degree 4 in LHS. Does this mean it has at most 4 optima? – Rittwik Chatterjee Dec 04 '22 at 01:07

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