Question on Increasing function

Question

Let $ f(x) = \frac{x^5}{5} + \frac{x^4}{4} + x^3 + \frac{kx^2}{2} + x$ be a real valued function. Then the greatest value of $k^2$ for which $f(x)$ is an increasing function $∀$ $x \in \mathbb{R}$

My approach was that :

$f'(x) = x^4+x^3+3x^2+kx+1$

$f''(x) = 4x^3+3x^2+6x+k$

$f'''(x) = 12x^2+6x+6 = 6(2x^2+x+1)$

ie, $f'''(x)$ has no real roots, and is always greater than zero

So, 1. the cubic $f''(x)$, will have only one root (ie. $f'(x)$ will have a minima or a maxima)

2. $f'(x)$ will always be greater than zero as we want the function to be increasing

Net, we can say that the graph of the $f'(x)$ will be upward opening, and will touch touch the $x-$ axis only at 1 point or always stay above it with one minima.

Can you please carry forward this method, or give some other solution, if possible, please try to refrain from giving solution completely based on graph making websites?

Answer 1

We have that the root $x_0$ for the cubic is attained for (it is a bijection since $f''(x)$ is strictly increasing)

$$f''(x) = 4x^3+3x^2+6x+k=0 \implies k_0=-4x_0^3-3x_0^2-6x_0$$

plugging this value in $f'(x)$ we obtain for the minimum the following value

$$f'(x_0) = -3x_0^4-2x_0^3-3x_0^2+1 $$

and we are interested in the value of $x_0$ such that $f'(x_0) =0$ with $k_0$ real which, by a numerical solution, leads to two different solutions

• $x_0\approx 0.46687798 \implies k_0 \approx -3.86226400$
• $x_0\approx -0.59074923\implies k_0 \approx 3.3221911$

therefore the maximum value for $k^2$ is $\approx 14.91708$.

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Here is a numerical check for the solution of $f'(x) = x^4+x^3+3x^2+k_0x+1$ which has indeed minimum equal to zero at $x=x_0$.