Understanding Curve Sketching for a Challenging Function

[kjland]

Hello everyone I'm having some difficulty wit curve sketching this specific function.
If some one could walk through the steps and solution so that i can hopefully grab a handle on the concept it would be much appreciated! I can handle the beginner functions, but this one is giving me a hard time, thank you.

Sketch the graph of
f(x) = x^3 / x^3 + 1

 

[MarkFL]

I am assuming we have:

$$f(x)=\frac{x^3}{x^3+1}$$

I would first look at $x$ and $y$ intercepts...what do you find?

Then I would focus on asymptotes...do you find any vertical/horizontal asymptotes?

If you are unsure how to find intercepts/asymptotes, let me know and I will explain. :)

 

[samir]

I am assuming we have:

$$f(x)=\frac{x^3}{x^3+1}$$

I would first look at $x$ and $y$ intercepts...what do you find?

Then I would focus on asymptotes...do you find any vertical/horizontal asymptotes?

If you are unsure how to find intercepts/asymptotes, let me know and I will explain. :)

$y$ intercept is at $x=0$.

$$x=0 \Leftrightarrow f(0)$$

$x$ intercept is at $y=0$.

$$y=0 \Leftrightarrow 0=\frac{x^3}{x^3+1}$$

$$f(0)=0$$

$$0=\frac{x^3}{x^3+1}$$

$$x=0$$

How do you find asymptotes?

This is a rational function. In other words the rule of the function is a rational expression. The dominator cannot be zero!

$$x^3+1 \neq 0$$

Argument? The function becomes undefined when the denominator is zero. Because dividing by zero breaks the universe!

But we can sort of defy the rule and make a new statement that makes the denominator equal to zero. This allows us to figure out at what x value the denominator becomes zero!

$$x^3+1 = 0$$

$$x=-1$$

Once we know this, we know at what x value the function is undefined. There will be a vertical asymptote at that x value (hence the name "vertical").

I still haven't learned how we find any horizontal asymptotes. Do we need the inverse function for this? Anyway! I don't think it's needed in this example.

So we have these points on the graph:

$$(0,0)$$

$$(0,0)$$

In other words the graph passes through these points. In fact, it's the same point!

We have to exclude $(-1,0)$ and all other $(-1,y)$ points, because this is where the vertical asymptote is.

Maybe you can complete this with a way to find the horizontal asymptote? (Smile)

For a nicer graph you can create a table of values, and then use these to approximate where the graph will pass.

 

[MarkFL]

For the horizontal asymptotes, we can write the function as:

$$f(x)=\frac{1}{1+\dfrac{1}{x^3}}$$ where $x\ne0$.

And then consider the limit:

$$\lim_{x\to\pm\infty}f(x)=1$$

So, we know $y=1$ is the horizontal asymptote.

If this has been posted in the Calculus forum, we could go on to examine slope and concavity for further clues about the function's behavior. However, the suggestion by samir to create and $xy$ chart is a good one to help find other points to give you an idea of how the graph behaves.