I have a question that tells me to find the range of $b^2+c^2$ for which $$f(x)= x^3+3x^2+4x+b \sin x + c \cos x,$$ $\forall x \in R$ is one-one function.
[The answer says that $b^2+c^2 \leq 1$]
My book does this question as follows:
$f'(x) = 3x^2+6x+4+b \cos x -c \sin x$
Now, the book states that the only possibility for $f(x)$ to be one-one would be if $f'(x) \geq 0$, i.e, it is monotonically increasing. I have a doubt here. Why cant $f'(x) \leq 0$, i.e, a monotonically decreasing function also satisfy the criteria that the function would be one-one?
Why can the function only be monotonically increasing and not decreasing for all $x$ in R??? I know that that the parabola is upwards, from $f'(x)$, but depending upon b and c values, can't the derivative of $f(x)$ be negative? Why is it said for one-one function the only possibility is that the function is increasing?
By C-S $$f'(x)=3x^2+6x+4+b\cos{x}-c\sin{x}\geq$$ $$\geq3(x+1)^2+1-\sqrt{(b^2+c^2)(\cos^2x+\sin^2x)}\geq1-\sqrt{b^2+c^2}.$$ We see that for $b^2+c^2\leq1$ we have $f(x)\geq0.$
From here easy to see that $b^2+c^2\leq1$ it's the condition for increasing.
$f$ can not be decreased because for any $a$ and $b$ we have: $$\lim\limits_{x\rightarrow+\infty}f(x)=+\infty$$ and $$\lim\limits_{x\rightarrow-\infty}f(x)=-\infty$$