I have come across a question online that I decided to have a stab at; it goes like this:


Let $f(x) := ax - \dfrac{x^3}{1 + x^2}$ where $a$ is a constant.

(i) Show that if $a \geq \dfrac{9}{8}$ then $f'(x) \geq 0\ \ \forall x$

(ii) Sketch the graph of $f$ according to the various value of $a$


For part (i), I showed this by first finding the derivative and then invoking the aforementioned condition on $a$:

$f'(x) = \dfrac{(a-1)x^4 + (2a - 3)x^2 + a}{(1 + x^2)^2} \geq \dfrac{(x^2 - 3)^2}{8(1 + x^2)^2} \geq 0\ \ \forall x$ as required.

Even though I think my demonstration is fine, I do have one question about this part. I did initially try to show the result by considering the discriminant of the numerator in $u = x^2$, which I denote by $D_u$ (because you cannot consider the typical quadratic discriminant for that fourth order equation, right?), according to the condition on $a$. However, I feel it became slightly confusing because: $D_u = -8a + 9 \geq 0$ according to the condition on $a$ given, which would mean that the quadratic in $u$ would have at least one real root. But then, it is only if that root (or those roots) is non-negative that we can determine roots for $f'(x)$. But then this discussion is pointless too because the equation can still have four roots yet never cross the $y$-axis. The only reason I bring this up is because the author seems to passingly suggest that considering the discriminant might help but he isn't clear as to whether this is contextual (that you can use it in other questions like this) or whether the notion pertains to this particular question - could somebody clarify this?

As for part (ii), the issue I have is in both cases, $a \geq \dfrac{9}{8}$ and $a < \dfrac{9}{8}$. In the first case, I know from part (i) that the graph is increasing (and also that $ f(0) = 0$) but the issue is I would have thought that the graph would take an indiscernible shape for larger values of $x$ since:

$f(x) = ax - \dfrac{x^3}{1 + x^2} = ax - \dfrac{1}{\frac{1}{x} + \frac{1}{x^2}}$

And so surely the right hand term would dominate this expression for larger $x$? However, when I looked in wolframalpha, the graph seems to be linear for larger values of $x$ - what gives? With regards to the second case, I don't know how to draw that because I don't know how to get information on $f'$ according to that case from the working I did in part (i).

Sorry for this rather drawn out discussion but I just wanted to make sure I put everything across.

Thank you in advance for any help you can give!


1 Answer 1


In order the derivative $$f'(x) = \dfrac{(a-1)x^4 + (2a - 3)x^2 + a}{(1 + x^2)^2}$$ be always positive, the numerator must not cancel for any value $x$ and moreover this would imply $a \gt 1$. So, as you did, setting $u=x^2$, you look for the roots of $$(a-1)u^2+(2a - 3)u + a=0$$ The discriminant is then $$\Delta=(2a-3)^3-4a(a-1)=9-8a$$ and it must be negative which then implies $a \gt \frac 98$. If $a=\frac 98$, $\Delta=0$, then a double root and, for this value $$f'(x) = \frac{\left(x^2-3\right)^2}{8 \left(x^2+1\right)^2}$$ which is always positive or zero as you did show.

Now, what happens when $x$ start to be very large ? The simplest way would be to say that $x^2+1\approx x^2$ and then $$f(x) = ax - \dfrac{x^3}{1 + x^2}\approx ax - \dfrac{x^3}{ x^2}=ax-x=(a-1)x$$ which is an oblique asymptote to the function. We could make better approximation using$$f(x) = ax - \dfrac{x^3}{1 + x^2}=ax-\dfrac{x}{1 + \frac 1{x^2}}$$ and remember that, for small values of $y$, $\frac 1{1+y}=1-y+y^2+\cdots$; replacing $y$ by $\frac 1{x^2}$ then gives $$f(x)\approx (a-1) x+\frac{1}{x}-\frac{1}{x^3}+\cdots$$ which shows the asymptote and also how it is approached for infinite values of $x$.

  • $\begingroup$ Thank you for that help; two questions, aren't you showing the reverse implication in the first part by implying that $a > \frac{9}{8}$? Ah ok so that is an oblique asymptote - but surely $f(0) = 0$ fine right? So how is the graph asymptotic? $\endgroup$
    – Necroticka
    Commented Jun 24, 2015 at 9:21
  • $\begingroup$ The graph is above the asymptote for large positive values of $x$ and below for large negative values of $x$ $\endgroup$ Commented Jun 24, 2015 at 9:26

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