Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 28

Use the guidelines of curve sketching to sketch the curve. $\displaystyle y = x^{\frac{5}{3}} - 5x^{\frac{2}{3}}$

The guidelines of Curve Sketching
A. Domain.
Since the function is continuous everywhere. It's domain is $(-\infty, \infty)$

B. Intercepts.
Solving for $y$-intercept, when $x= 0$
$y = (0)^{\frac{5}{3}} - 5(0)^{\frac{2}{3}} = 0$
Solving for $x$-intercept, when $y = 0$

$
\begin{equation}
\begin{aligned}
0 & = x^{\frac{5}{3}} - 5x^{\frac{2}{3}}\\
\\
5x^{\frac{2}{3}} &= x^{\frac{5}{3}}\\
\\
5 &= x^{(5/3 - 2/3)}\\
\\
x &= 5
\end{aligned}
\end{equation}
$



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

D. Asymptotes.
The function has no asymptotes.

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

$
\begin{equation}
\begin{aligned}
f'(x) &= \frac{5}{3} x^{\frac{2}{3}} - 5 \left( \frac{2}{3} x^{\frac{-1}{3}} \right)\\
\\
f'(x) &= \frac{5}{3} \left[ x^{\frac{2}{3}} - 2x^{\frac{-1}{3}} \right]
\end{aligned}
\end{equation}
$


when $f'(x) = 0$

$
\begin{equation}
\begin{aligned}
0 &= x^{\frac{2}{3}} - 2x^{\frac{-1}{3}}\\
\\
0 &= x \left( x^{\frac{-1}{3}} - 2x^{\frac{-4}{3}} \right)
\end{aligned}
\end{equation}
$

We have, $x=0$ and $x^{\frac{-1}{3}} - 2x^{\frac{-4}{3}} = 0$

Therefore, the critical numbers $x = 2$ and $x = 0$. Hence, the interval of increase or decrease is.

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
x < 0 & + & \text{increasing on } (-\infty, 0)\\
\hline\\
0 < x < 2 & - & \text{decreasing on } (0,2)\\
\hline\\
x > 2 & + & \text{increasing on } (2,\infty)\\
\hline
\end{array}
$


F. Local Maximum and Minimum Values.
Since $f'(x)$ changes from negative to positive at $x=2$. Thus $f(2) = -4.76$ is a local minimum. On the other hand, since $f'(x)$ changes from positive to negative at $x = 0$, thus, $f(0) = 0$ is a local maximum.

G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= \frac{5}{3} \left[ x^{\frac{2}{3}} - 2x^{\frac{-1}{3}} \right], \text{ then}\\
\\
f''(x) &= \frac{5}{3} \left[ \frac{2}{3} (x)^{\frac{-1}{3}} -2 \left( \frac{-1}{3}\right) (x)^{\frac{4}{3}} \right]\\
\\
f''(x) &= \frac{10}{9} x^{\frac{-1}{3}} + \frac{10}{9} x^{\frac{-4}{3}}
\end{aligned}
\end{equation}
$


when $f'(x) = 0$

$
\begin{equation}
\begin{aligned}
0 &= \frac{10}{9} \left( x^{\frac{-1}{3}} + x^{\frac{-4}{3}} \right)\\
\\
0 &= x^{\frac{-1}{3}} + x^{\frac{-4}{3}}\\
\\
0 &= x \left( x^{\frac{-4}{3}} + x^{\frac{-7}{3}} \right)
\end{aligned}
\end{equation}
$

We have $x = 0$, as our inflection points
Hence, the concavity is...

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


H. Sketch the Graph.