Domain of k in $f(x)=x^3+kx^2+5x+4\sin^2x$

Question

The question says:

Let $f(x)=x^3+kx^2+5x+4\sin^2x$ be an increasing function on $x \in R$. Then domain of k is?

This is What I tried:

I tried differentiating the expression but $f'(x)$ didn't turn out to be a quadratic(as it has that $\sin2x$ term) $$f'(x)=3x^2+2kx+5+4\sin2x>0$$

So how else am I supposed to solve this question?

Answer 1

The set up of the problem is completely right with only one remark: you can set $$f'(x)\ge0$$ to have $f(x)$ increasing and maybe also strictly increasing since the the set on which $f'(x)=0$ for "regular functions" like the given doesn't contain any intervals.

You can refer also to this other answer for a detailed comment on this latter fact.

For a first estimation of the range note that for

$$3x^2+2kx+5\ge4$$

the condition is satisfy, thus from here we can find a first range for $k\in[-\sqrt 3,\sqrt 3]$, indeed

$$3x^2+2kx+1\ge0 \iff \Delta=4k^2-12\le 0$$

To improve the result we could try to find the minimum of $f'(x)$ and set it to $0$, that is

$$f''(x)=6x+2k+8\cos2x=0 \implies x=x_k$$

and

$$f'(x_k)=0$$

Answer 2

From $f '(x)=3x^2+2kx+5+4\sin2x$, try to solve $f'(x)=0$, so $$k=\frac{-3x^2-5-4\sin2x}{2x}$$ If you can't solve $f'(x)=0$, then $f(x)$ is increasing, so the domain of $k$ is everything outside the range of that function.