Graphing $y= \frac{x^3}{x^2-1}$

Question

I'm having a lot of problems trying to graph this... other than using a graphics calculator!

I know the domain is all real values of $x , x$ does not equal $-1 \text{or} 1$. The point of inflection is easy to find since when $x=0 , y=0.$ How do I find the local maxima and minima?

Answer 1

Let $f(x):=\displaystyle\frac{x^3}{x^2-1}$. $\ $ Find out the following informations:

1. $\displaystyle\lim_{x\to-1-0} f(x)$, $\ \ \ \displaystyle\lim_{x\to-1+0} f(x)$
2. $\displaystyle\lim_{x\to +1-0} f(x)$, $\ \ \ \displaystyle\lim_{x\to +1+0} f(x)$
3. $\displaystyle\lim_{x\to-\infty} f(x)$, $\ \ \ \displaystyle\lim_{x\to +\infty}f(x)$
4. $f'(x)$, $\ \ f''(x)$
5. where $f'(x)=0$ -- these might be the local maximum/minimum points. (Because if a differentiable function has local maximum or minimum at $x$, then its tangent line is horizontal there.)
6. if $f'(x)>0$, then $f$ is increasing locally around $x$
7. if $f''(x)>0$ then $f$ is convex locally around $x$ (convex from upwards, i.e. smiley-like)
8. if $x$ is an inflection point (where $f$ is changing convexity), we necessarily have $f''(x)=0$.

Answer 2

Even without calculus, we can make a rough sketch. $$y=\frac{x^3}{(x+1)(x-1)}$$

This function is odd, so it's rotationally symmetric about the origin. It is positive on $(1,+\infty)$, negative on $(0,1)$, positive on $(-1,0)$, and negative on $(-\infty, -1)$. It blows up as it approaches $\pm 1$, so $\lim_{x\rightarrow 1+}y=+\infty$, while $\lim_{x\rightarrow 1-}y=-\infty$. Similarly $\lim_{x\rightarrow -1+}y=+\infty$, while $\lim_{x\rightarrow -1-}y=-\infty$. The degree of the numerator is greater than the degree of the denominator (by 1), so $\lim_{x\rightarrow \infty}y=\infty$, and $\lim_{x\rightarrow -\infty}y=-\infty$. Hence one should expect a local minimum somewhere in $(1,\infty)$ and a local maximum somewhere in $(-\infty,-1)$. The only place it crosses the $x$-axis is where the numerator is zero, namely $x=0$.

To compute exactly what the minima are, and to determine inflection points, you'll need to take the derivative though. Note that $x=0$ may or may not be an inflection point; the correct test is given by @Berci's answer.