How to sketch this function/find the minima/identify behaviour

25 Views Asked by At

I need to sketch this function: $f(x) = \frac{x(1-2x^2)}{2(1+x)}$. I've found the zeros - at $y=0$, $x=0,\frac{\sqrt 2}{2}$ and $-\frac{\sqrt 2}{2}$. I also found the infinities - as $x \to +\infty, y \to -\infty$ and as $x \to -\infty, y \to +\infty$. I tried to find the stationary points by finding $f'(x)=\frac{-4x^3-6x^2+1}{2(x+1)^2}$ and $f''(x)=\frac{-2x^3-6x^2-6x-1}{(x+1)^3}$, but I can't solve the numerators=0 analytically using the factor theorem, which made it all seem a bit pointless. I put the function into wolfram alpha and saw that it behaves like a cubic, but when looking at the original function, I would have guessed it behaves more like a quadratic. I was wondering how I could have inferred that it behaves like a cubic? Also just any general tips on how I could sketch this function in the first place would also be welcome. Thanks!

1

There are 1 best solutions below

0
On BEST ANSWER

You're doing well so far.

but I can't solve the numerators=0 analytically using the factor theorem, which made it all seem a bit pointless.

You won't find integer roots, but there is a rational root which you could find using the rational root theorem. Once you find that $-\tfrac{1}{2}$ is a root, you can factor and solve further: $$\begin{align}-4x^3-6x^2+1=0 & \iff-\left(2x+1\right)\left(2x^2+2x-1\right)=0 \\[6pt] & \iff x=-\tfrac{1}{2} \;\vee\;x=\tfrac{-1\pm\sqrt{3}}{2} \end{align}$$ Check the sign changes of $f'$ or look at the sign of $f''$ to determine where you have a minimum/maximum.

For $f''(x)$, note that you can rewrite the numerator: $$-2x^3-6x^2-6x-1=1-2\left(x^3+3x^2+3x+1\right)=1-2\left(x+1\right)^3$$ and you find the (only) root at $x=\tfrac{1}{\sqrt[3]{2}}-1$.

and saw that it behaves like a cubic, but when looking at the original function, I would have guessed it behaves more like a quadratic.

I would agree that for large $x$, you expect a quadratic-like behavior as for large $x$ we have: $$\frac{x(1-2x^2)}{2(1+x)}\approx\frac{1-2x^2}{2}$$

Also just any general tips on how I could sketch this function in the first place would also be welcome.

Combine all the info you have and make sign tables of $f$, $f'$ (increasing/decreasing) and $f''$ (convex/concave). Note that there is a vertical asymptote at $x=-1$; investigate the function's behavior there too (left and right limits).