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If $f(x)=x^3+3x^2+4x+ a \sin x + b\cos x ~ \forall x \in \mathbb{R}$ is an injection then the greatest value of $a^2+b^2$ is _______?

To ensure injection, we must ensure that there is no maxima/minima in any interval which is equivalent to $f'(x)\neq 0$.

Note that $f'(x)=3x^2+6x+4+a \cos x - b \sin x \neq 0$. It can be observed that $3x^2+6x+4>0$ with its minimum value being $1$.

So, our condition can be reduced to $a \cos x - b \sin x > -1$.

Again, $a \cos x - b \sin x +1$ can be written as $\frac{a}{\sqrt {a^2+b^2}} \cos (-x) - \frac{b}{\sqrt{a^2+b^2}} \sin (-x) +1$ such that $\sin (\theta -x) + \frac{1}{\sqrt {a^2+b^2}} >0$.

I couldn't proceed anymore!

DeepSea
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Mathejunior
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1 Answers1

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Introducing $t=x+1$, rewrite the requirement on $f'(x)$ as: $$ 3t^2+1+A\cos(t+\phi) \ge 0. $$ To maximize the value of $A=\sqrt{a^2+b^2}$ the expression should attend its minimal value as far from $t=0$ as possible, which corresponds to the choice $\phi=0$.

In this case for sufficiently large $A$ two symmetric minima on both sides of $t=0$ will appear. If $A$ is increased further the minima become negative.

The critical values of the parameters $A$ and $t$ correspond to the case when the function value in the minima is 0: $$\begin{cases} 3t^2+1+A\cos t&=0\\ 6t-A\sin t&=0 \end{cases}. $$ For the critical value of $t$ one obtains then the equation: $$ (3t^2+1)\sin t+6t\cos t=0, $$ which minimal positive solution is: $$ t_*\approx2.49037. $$

The corresponding value of $A$ is: $$ A_*=\frac{6t_*}{\sin t_*}\approx24.6508. $$

Its square gives the greatest possible value of $a^2+b^2$.

user
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