Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 46

a.) Use a graph of $f(x) = x^3(x-2)^4$ to give a rough estimate of the intervals of concavity and the coordinates of the points of inflection.
b.) Use a graph of $f''$ to give better estimates.


a.)


Based from the graph, the function has an upward concavity at intervals, $0 < x < 0.4$ and $ x > 1.3$. On the other hand, the funciton has a downward concavity at interval $x < 0 $ and $0.4 < x < 1.3$. Also the coordinates of the points of inflections can be approximated as $(0,0),(0.4,0.5)$ and $(1.3,0.5)$


$
\begin{equation}
\begin{aligned}
\text{if } f (x) &= x^3(x-2)^4, \text{ then}\\
\\
f'(x) &= x^3 \left[ 4 (x-2)^3 \right] + 3x^2(x-2)^4 && \Longleftarrow \text{(By using Chain and Product Rule)}\\
\\
f''(x) &= x^3 \left[ 12(x - 2)^2 \right] + 3x^2 \left( 4(x-2)^3\right) + 3x^2 \left[4 (x-2)^3 \right]+ 6x (x-2)^4\\
\\
f''(x) &= 12x^3 (x-2)^2 + 12x^2 (x-2)^3 + 12x^2 (x-2)^3 + 6x(x-2)^4\\
\\
f''(x) &= 12x^3(x-2)^2 + 24 x^2 (x-2)^3 + 6x (x-2)^4
\end{aligned}
\end{equation}
$





Based from the graph of $f''$, the function has an upward concavity (where $f''$ is positive) at intervals $0 < x < 0.46$ and $x > 1.28$ On the other hand, the function has downward concavity (where $f''$ is negative) at intervals $x < 0 $ and $ 0.46 < x < 1.28$
Also using the graph of $f''$, the points of inflections (where the slope is 0) are approximately $x = 0.18, \quad x = 0.83, \quad x = 1.57, \quad x = 2$