Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 10

Suppose that $4x^3 + 3x^2 - 6x + 1$

a.) Determine the intervals on which $f$ is increasing or decreasing.

If $f(x) = 4x^3 + 3x^2 - 6x + 1$, then


$
\begin{equation}
\begin{aligned}

f'(x) =& 12x^2 + 6x - 6
\\
\\
f''(x) =& 24x + 6

\end{aligned}
\end{equation}
$


To find the critical numbers, we set $f'(x) = 0$. So..


$
\begin{equation}
\begin{aligned}

f'(x) = 0 =& 12x^2 + 6x - 6
\\
\\
0 =& 12x^2 + 6x - 6

\end{aligned}
\end{equation}
$


By using Quadratic Formula, we obtain the critical numbers as,

$\qquad \displaystyle x = \frac{1}{2}$ and $x = -1$

Hence, we can divide the intervals of $f$ by:

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f \\
\hline\\
x < -1 & + & \text{increasing on} (- \infty, -1) \\
\hline\\
\displaystyle -1 < x < \frac{1}{2} & - & \displaystyle \text{decreasing on} \left( -1, \frac{1}{2} \right) \\
\hline\\
\displaystyle x > \frac{1}{2} & + & \displaystyle \text{increasing on} \left( \frac{1}{2} , \infty \right)\\
\hline
\end{array}
$

These values are obtained by evaluating $f''(x)$ within the specified
interval. The concavity is upward when the sign of $f''(x)$ is positive. On the
other hand, the concavity is downward when the sign of $f''(x)$ is negative.

b.) Find the local maximum and minimum values of $f$.

We will use the Second Derivative Test to evaluate $f''(x)$ at these critical numbers:


$
\begin{equation}
\begin{aligned}

& \text{So when } x = \frac{1}{2},
&& \text{when} x = -1
\\
\\
& f'' \left( \frac{1}{2} \right) = 24 \left( \frac{1}{2} \right) + 6
&& f''(-1) = 24 (-1) + 6
\\
\\
& f''\left( \frac{1}{2} \right) = 18
&& f''(-1) = -18

\end{aligned}
\end{equation}
$





Since $\displaystyle f' \left( \frac{1}{2} \right) = 0$ and $\displaystyle f''\left( \frac{1}{2} \right) > 0, f \left( \frac{1}{2} \right) = \frac{-3}{4}$ is a local minimum.

On the other hand, since $f'(-1) = 0$ and $f''(-1) < 0, f(-1) = 6$ is a local maximum.


c.) Find the intervals of concavity and the inflection points.

We set $f''(x) = 0$ to determine the inflection points..


$
\begin{equation}
\begin{aligned}

0 =& 24x + 6
\\
\\
x =& \frac{-6}{24}

\end{aligned}
\end{equation}
$


The inflection point is at $\displaystyle x = - \frac{1}{4}$

Let's divide the interval to determine the concavity

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity} \\
\hline\\
x < \frac{-1}{4} & - & \text{Downward} \\
\hline\\
x > \frac{-1}{4} & + & \text{Upward}\\
\hline
\end{array}
$

These data obtained by substituting any values of $x$ to $f'(x)$ within the specified interval. Check its sign, if it's positive, it means that the curve is increasing on that interval. On the other hand, if the sign is negative, it means that the curve is decreasing on that interval.