Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 30

Use the guidelines of curve sketching to sketch the curve. $\displaystyle y = \sqrt[3]{x^3+1}$

The guidelines of Curve Sketching
A. Domain.
We know that cube root functions are defined for all values of $x$. Thus, the domain $(-\infty, \infty)$

B. Intercepts.
Solving for $y$-intercept, when $x = 0$
$y = \sqrt[3]{(0)^3 + 1} = 1$
Solving for $x$-intercept, when $y = 0$

$
\begin{equation}
\begin{aligned}
0 & = \sqrt[3]{x^3 +1}\\
\\
x^3 &= -1\\
\\
x &= -1
\end{aligned}
\end{equation}
$



C. Symmetry.
The function is not symmetric to either $y$-axis or origin by using symmetry test.

D. Asymptotes.
There are no vertical asymptotes as well as horizontal asymptotes since $\displaystyle \lim_{x \to \pm \infty} \sqrt[3]{x^3 + 1} = \pm \infty$.

E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$,

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{1}{3} (x^3 + 1)^{\frac{-2}{3}} (3x^2)\\
\\
f'(x) &= \frac{x^2}{(x^3+1)^{\frac{2}{3}}}
\end{aligned}
\end{equation}
$


Notice that $f'(x) > 0$ for all except $x = 0$ and $x = 1$ where $f(x)$ is not differentiable. Therefore, $f(x)$ is increasing on the entire domain.

F. Local Maximum and Minimum Values.
Since $f(x)$ is always increasing, therefore, the function has no local maxima or minima.

G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= \frac{x^2}{(x^3+1)^{\frac{2}{3}}} \text{, then}\\
\\
f''(x) &= \frac{(x^3+1)^{\frac{2}{3}} (2x) - x^2 \left(\frac{2}{3}(x^3+1)^{\frac{-1}{3}} (3x^2) \right) }{\left[ (x^3+1)^{\frac{2}{3}} \right]^2}
\end{aligned}
\end{equation}
$

Which can be simplied as....
$\displaystyle f''(x) = \frac{2x}{(x^3+1)^{\frac{5}{3}}}$

If we divide the interval to determine the concavity...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity}\\
\hline\\
x < -1 & + & \text{Upward}\\
\hline\\
-1 < x < 0 & - & \text{Downward}\\
\hline\\
0 < x < 1 & + & \text{Upward}\\
\hline\\
x > 1 & - & \text{Downward}\\
\hline
\end{array}
$

Based from these data, we can say that the function has inflection points on $f(-1) =0$, $f(0) = 1$, and $f(1) = \sqrt[3]{2}$

H. Sketch the Graph.