Condition for function to increase

Question

If $f(x) = x^5/5 + x^4/4 + x^3 + kx^2/2 + x$ is an increasing function, for all $x$, then $k^2$ can be?

1. 3
2. 5
3. 7
4. 13

The derivative of this function is $x^4 + x^3 + 3x^2 + (k)x + 1$, which we have to make positive. Factoring out $x^2$, we get g(x)=$ x^2(x^2 + \frac{1}{x^2} +2 +1+ x + \frac{k}{x})=x^2((x+\frac{1}{x})^2+1+(x +\frac{k}{x})$

Now, $x^2$ is positive , and if we analyse the function inside the bracket, $(x+\frac{1}{x})^2+1$ is always $\ge5$. Therefore, if we make the minimum value of the remainig expression, i.e $(x +\frac{k}{x})$, greater than $(-5)$, we will have $g(x)=f'(x) \ge0$. Since the minimum value is $-2\sqrt(k)$, we get $k<=25/4$. Which means all options 1,2,3,4 must be correct.

However,the correct answers are 1,2,3 only. And indeed, wolfram alpha confirms that for K=$\sqrt(13)$, the derivative becomes -ve in an interval.

Where have I gone wrong?

Answer 1

I think you are right!

$$f'(x)=x^4+x^3+3x^2+kx+1.$$ We see that for any $k^2\in\{3, 5,7\}$, we obtain:

$$f'(x)=x^4+x^3+3x^2+3x+1=x^4+x^3+\frac{1}{4}x^2+\frac{11}{4}x^2+kx+1>0$$ because $$k^2-4\cdot\frac{11}{4}\cdot1<0.$$

For $k=\sqrt{13}$ we see that $f'\left(-\frac{1}{2}\right)=\frac{1}{16}(27-8\sqrt{13})<0$, but for $k=-\sqrt{13}$ easy to show that $f'(x)>0$, which says that $k^2$ can be any number from the given.

We obtain: $$x^4+x^3+3x^2-\sqrt{13}x+1=\left(x^2+\frac{1}{2}x-\frac{1}{2}\right)^2+\frac{1}{4}(15x^2-2(2\sqrt{13}-1)+3)>0$$ because $$(2\sqrt{13}-1)^2-15\cdot3<0.$$

Answer 2

The troubles in your reasoning is that with $k >0$ at $x=-\sqrt{k}$ you have a local maximum and not a minimum for the function $f(x)=x+\frac{k}{x}$. If $k<0$ you don't have any minimum.

On the other side if $k=\sqrt{13}$ you have that $f'(-1)>0$ while $f'(-1/2)>0$ thus you have a zero of $f'(x)$ in $(-1,-1/2)$. Moreover $f'(0)=1>0$ thus you have another zero of $f'(x)$ in $-1/2,0)$. This prove that for such $k=\sqrt{13}$ your function is not monotone. (below the graph of $f'(x)$ for $k=\sqrt{13}$