Sketch the graph of
\[ y = \frac{1}{x^3 + 1} \]
Sketch the inner function
The inner function is a standard cubic shifted one unit in the positive y direction.
Find intercepts of 1/f:
When x = 0, 1/(x3 + 1) = 1
When x3 + 1 = 0, x = -1 so 1/(x3 + 1) is not defined at x = -1
Find the asymptotes
When x is large and positive, 1/(x3 + 1) approaches zero
When x is large and negative, 1/(x3 + 1) approaches zero
So y = o is a horizontal asymptote.
From the intercept analysis, x = -1 is a vertical asymptote
To determine where the two graphs intersect, solve
\[ x^3 + 1 = \frac{1}{x^3 + 1} \]
This gives
\[ \begin{align}
(x^3 + 1)^2 &= 1 \\
x^3 + 1 &= \pm 1 \\
x^3 = 0 \quad &\text{and} \quad x^3 = -2
\end{align} \ \]
so the intersection point is x = 0, y = 1 and x = -1.26, y = -1.
Investigate the point x = 0
When x = -1/2, y = 8/7. When x = -1/4, y = 64/63.
When x = 1/2, y = 8/9. When x = 1/4, y = 64/65.
So the curves becomes horizontal at x = 0.
Your sketch should look something like this:
Important This is not a 'textbook on a screen' - this is an interactive learning system. Which means you have to 'act' to make it go.
Look through the options below and select the one you think is the best response. You will gain most benefit if you work the problem with pen and paper as you go, i.e. decide what you think is the correct option and write down what you think the result will be before you click. If the next step doesn't display, click another option.