If $f(x)=x^3+x^2+ax+4\sin x$ is increasing for all real values of $x$, find the interval in which the variable $a$ lies.

Question

I have this problem

If $f(x)=x^3+x^2+ax+4\sin x$ is increasing for all real values of $x$, then $a$ lies in the interval

So what I did was to first differentiate the function and equate it to greater than equal to zero ie

$f'(x)=3x^2+2x+a+4\cos x \geq 0$

I also know that the determinant of this value should be less than 0.
But I don't know what to do after this. I asked my teacher and some other sources and they just substituted $\cos x=-1$ and wrote the equation as :
$f'(x)=3x^2+2x+a-4 \geq 0$

I don't know what to do after this and why did my teacher do so. If you can guide me then it would be great.

Answer 1

To show why the idea of "just replace $\cos(x)$ with it's smallest value $-1$" doesn't give you the whole range of $a$, look at the following picture, that plots both the actual derivative $f'(x)=3x^2+2x+a+4\cos(x)$ in red and the "replacement" $r(x)=3x^2+2x+a-4$ in black, for $a=0$:

$r(x)$ is negative for some $x$ around $0$, and that's why Devansh Kamra's solution says $a$ should be at least $\frac{13}3$. But as can be seen, the actual derivative (in red) is not negative anywhere. So $a=0$ is a valid choice for $a$, that Devansh Kamra's solution ignores.

The reason this is so is that "just replace $\cos(x)$ with it's smallest value $-1$" is conservative: You can see in the picture that the red plot is always above the black plot, which is easy to understand because the difference between those 2 functions is $4\cos(x)+4$, which is $\ge 0$. Devansh Kamra's solution correctly finds exactly all $a$ where the replacement function $r(x)$ is non-negative. This in turn makes sure that the actual derivative $f'(x)$ is also non-negative, so we are now sure that for all $a \ge \frac{13}3$ $f'(x)$ is non-negative.

But we can't make the reverse conclusion, we don't know if more $a$ exist. Because we assumed "the worst case" for $\cos(x)$ (that it is $-1$), we are missing $a=0$, where around $x=0$ $\cos(x)$ is near $1$, so the $+4\cos(x)$-term is "actively helping" to keep $f'(x)$ above $0$.

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So how to find the actual range of $a$ such that $f'(x) \ge 0$?

We have $f'(x)=3x^2+2x+a+4\cos(x)$, so $a$ is just an added constant.

If we find the minumum value of $g(x)=3x^2+2x+4\cos(x)$, we have found the maximal possible $-a$.

So we have $g'(x)=6x+2-4\sin(x), g''(x)=6-4\cos(x)>0$. $g(x)$ is dominated by the quadratic term $3x^2$ for large $|x|$ and $g''$ is always positive, so solving $g'(x)=0$ will find the $x$-value where $g(x)$ becomes minimal.

As can be seen here, there is exactly one solution:

I don't think there is a way to find the solution explicitely, a common fix-point iteration would be

$$x_0=-0.8, x_{n+1}=\frac{2\sin(x_n)-1}3$$

and this leads to the solution $x_s \approx -0.82140...$ and $g(x_s)\approx 3.106...$

This means that the interval for $a$ where $f'(x)$ is always non-negative is

$$a \ge -g(x_s) \approx -3.106...,$$

where $x_s$ is the only real solution to $6x+2=4\sin(x)$, $x_s \approx -0.82140...$.

Answer 2

Note that the minimum value of the cosine term is $-1.$ Thus, if the function $f'$ is to be positive for all real $x,$ then we must consider the least value the function can take, and choose $a$ so that this value is positive. Hence the setting of $\cos x=-1.$

Answer 3

Note that $f'(x)=3x^2+2x+a+4\cos x$ consists of two functions, $g(x)=3x^2+2x+a$ and $h(x)=4\cos x$ $$g'(x)= 6x+2$$ equating that to $0$, we get: $$x=\dfrac{-1}{3}$$ which means the minimum value of $g(x)$ will be at $x=\dfrac{-1}{3}$ $$g_{min}(x)=g(\dfrac{-1}{3})=\dfrac{3a-1}{3}$$ The minimum value of $h(x)$(for any value of $x$, not necessarily $\dfrac{-1}{3}$) is $-4$

$\therefore$ if $\dfrac{3a-1}{3}-4\geq 0$, then for all values of $x$, $f'(x)\geq 0$ $$\dfrac{3a-1}{3}-4\geq 0$$ $$a\geq \dfrac{13}{3}$$