If $f(x)=\frac{x^5}5+\frac{x^4}4+x^3+\frac{kx^2}2+x$ be a real valued function. Find the maximum value of $k^2$ for which $f(x)$ is increasing $\forall x \in R$.
This question already has answers here, but I am wondering if there is a solution that doesn't find $k$ numerically or using discriminant of quartic.
Maybe something along the following attempt?
$f'(x)=x^4+x^3+3x^2+kx+1$
$f''(x)=4x^3+3x^2+6x+k$
$f'''(x)=12x^2+6x+6$
Discriminant of quadratic is negative. So, cubic has only one root.
It means quartic has only one local minima, and no local maxima.
Now that minima should not be below x-axis, for $f(x)$ to be increasing.
Say, quartic minima occurs at $x=\alpha$.
So, $f'(\alpha)\ge0$ and $f''(\alpha)=0$.
But don't know how to conclude.
Below is a natural solution which goes very smooth but to the last step.
The function $f=f(x,k)$ increases on $\mathbb R$ iff $f’(x,k)\ge 0$ for each $x\in\mathbb R$.
Put $K=\{k\in\mathbb R: f'(x,k)\ge 0$ for each $x\in\mathbb R\}$.
Pick any $k\in K$.
Then $6+k=f’(1,k)\ge 0$ and $4-k=f’(-1,k)\ge 0$, so $k\in [-6,4]$.
Thus $K$ is bounded.
It is easy to check that $f’(x,k)>0$ for each $x\in\mathbb R\setminus [-5,5]$ and each $k\in [-10,10]$.
Suppose that $k\in\mathbb R\setminus K$.
Then there exists $x\in X$ such that $f’(x,k)<0$.
Since the function $f’(x,k)$ is continuous with respect to $k$, there exists $\delta>0$ such that $f’(x,k’)<0$ for each $k’\in (k-\delta, k+\delta)$.
So $\mathbb R\setminus K$ is open and thus $K$ is closed.
Let $k$ be any boundary point of $K$.
Suppose for a contradiction that $f’(x,k)>0$ for each $x\in\mathbb R$.
Since the segment $[-5,5]$ is compact, the set $f’([-5,5],k)$ is closed and so there exists $\varepsilon\in (0,1)$ such that $f’(x,k)>\varepsilon$ for each $x\in [-5,5]$.
Let $k’$ be any number from $(k-\varepsilon/5,k+\varepsilon/5)$.
Pick any $x\in [-5,5]$.
Then $|f’(x,k’)- f’(x,k)|= |k’x-kx|=|k’-k||x|<\varepsilon$, so $f’(x,k’)> f’(x,k)-\varepsilon>0$.
Since $f’(x,k’’)>0$ for each $x\in\mathbb R\setminus [-5,5]$ and each $k’’\in [-10,10]$, we have that $f’(x,k’)>0$ for each $x\in\mathbb R$.
That is, $k’\in K$ for any $k'\in (k-\varepsilon/5,k+\varepsilon/5)$, which is impossible since $k$ is a boundary point of $K$.
Thus $x^4+x^3+3x^2+kx+1=f’(x,k)=0$ for some $x\in\mathbb R$.
Therefore $f’(x,k)$ attains a minimum at $x$, so $4x^3+3x^2+6x+k=f’’(x,k)=0$.
The obtained system can be solved for $x$ and (all) boundary points $k$ of $K$.
In particular, we have $-kx=x^4+x^3+3x^2+1=4x^4+3x^3+6x^2$, and thus $3x^4+2x^3+3x^2-1=0$.
The latter equation has the fourth degree, so it can be solved in radicals.
Unfortunately, when I tried to find the roots of the equation by Mathcad, I obtained the message that the result is too long to display.
The numerical solution provides two real roots $x$, $\approx -0.590749$ and $\approx 0.466878$, and the respective values of $k$, $\approx 3.32219$ and $\approx -3.862264$.