Single Variable Calculus, Chapter 4, 4.6, Section 4.6, Problem 2

Use the graphs of $f'$ and $f''$ to estimate the intervals of increase and decrease, extreme values, intervals of concavity and inflection points. Suppose that $f(x) = x^6 - 15x^5 + 75x^4 -125x^3 - x$.

If $f(x)= x^6 - 15x^5 + 75x^4 - 125x^3 - x$, then


$
\begin{equation}
\begin{aligned}

f'(x) =& 6x^5 - 75x^4 + 300x^3 - 375x^2 - 1
\\
\\
f''(x) =& 30x^4 - 300x^3 + 900x^2 - 750x

\end{aligned}
\end{equation}
$








Based from the graph of $f', f'(x) > 0$ ($f$ is increasing) on the interval $x > 2.5$. On the other hand, $f'(x) < 0$ and $f$ is decreasing on the interval $x < 2.5$.

Based from the graph of $f$, we can estimate the value of local minimum (where the slope is 0) a $f(2.5) \approx -250$.

Based from the graph of $f''$, the intervals where the function has downward concavity are $0 < x < 1.3$ and $3.7 < x < 5$. On the other hand, the intervals where the function has upward concavity are from $x < 0, 1.3 < x < 3.7$ and $x > 5$. Therefore, the points of inflection can be approximated as $f(0) \approx 0, f(1.3) \approx -110, f(3.7) \approx -115$ and $f(5) \approx -5$.