# Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 34

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Suppose that $f(x) = x^5 - 2x^3 + x$

a.) Determine the intervals of increase or decrease.

If $f(x) = x^5 - 2x^3 + x$ then,

\begin{aligned} f'(x) =& 5x^4 - 6x^2 + 1 \\ \\ f''(x) =& 20x^3 - 12x \end{aligned}

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

\begin{aligned} f'(x) = 0 =& 5x^4 - 6x^2 + 1 \\ \\ 0 =& 5x^4 - 6x^2 + 1 \\ \\ 0 =& (5x^2 - 1)(x^2 - 1) \end{aligned}

The critical numbers are at

$\begin{array}{ccc} \displaystyle x = \pm \sqrt{\frac{1}{5}} & \text{and} & x = \pm \sqrt{1} \\ \displaystyle x = \pm \frac{1}{\sqrt{5}} & \text{and} & x = \pm 1 \end{array}$

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

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f'(x) & f \\ \hline\\ x < -1 & + & \text{increasing on} (- \infty, -1) \\ \hline\\ -1 < x < \frac{-1}{\sqrt{5}} & - & \text{decreasing on} \left( - 1, \frac{-1}{\sqrt{5}} \right) \\ \hline\\ \frac{-1}{\sqrt{5}} < x < \frac{1}{\sqrt{5}} & + & \text{increasing on} \left( \frac{-1}{\sqrt{5}} , \frac{1}{\sqrt{5}}\right) \\ \hline\\ \frac{1}{\sqrt{5}} < x < 1 & - & \text{decreasing on} \left( \frac{1}{\sqrt{5}}, 1 \right) \\ \hline\\ x > 1 & + & \text{increasing on} (1, \infty)\\ \hline \end{array}$

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.

b.) Find the local maximum and minimum values.

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

\begin{aligned} & \text{So when x = 1,} && \text{when x = -1,} \\ \\ \\ & f''(1) = 20(1)^3 - 12(1) && f''(-1) = 20(-1)^3 - 12 (-1)\\ \\ \\ & f''(1) = 8 && f''(-1) = -8 \end{aligned}

\begin{aligned} \text{when } x =& \frac{1}{\sqrt{5}}, && \text{when } x = \frac{-1}{\sqrt{5}} \\ \\ f'' \left( \frac{1}{\sqrt{5}} \right) =& 20 \left( \frac{1}{\sqrt{5}} \right)^3 - 12 \left( \frac{1}{\sqrt{5}} \right) && f'' \left( \frac{-1}{\sqrt{5}} \right) = \frac{8 \sqrt{5}}{5} \end{aligned}

Since $f'(1)$ and $\displaystyle f'\left( \frac{-1}{\sqrt{5}} \right) = 0, f''(1)$ and $\displaystyle f''\left( \frac{-1}{\sqrt{5}} \right) > 0$ are local minimums. On the other hand, since $f(-1)$ and $\displaystyle f' \left( \frac{1}{\sqrt{5}} \right) = 0, f''(-1)$ and $\displaystyle f''\left( \frac{1}{\sqrt{5}} \right) < 0, f(-1) = 0$ and $\displaystyle f\left( \frac{1}{\sqrt{5}} \right) = 0.2862$ are local maximums.

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

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

\begin{aligned} f''(x) = 0 =& 20x^3 - 12x \\ \\ 0 =& 20x ^3 - 12x \\ \\ 0 =& 4x (5x^2 - 3) \\ \\ x =& 0 \text{ and } 5x^2 - 3 = 0 \end{aligned}

Therefore, the inflection points are $x = 0$ and $\displaystyle x = \pm \sqrt{\frac{3}{5}}$.

Let's divide the interval to determine the concavity..

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f''(x) & \text{Concavity} \\ \hline\\ \displaystyle x < - \sqrt{\frac{3}{5}} & - & \text{Downward} \\ \hline\\ \displaystyle - \sqrt{\frac{3}{5}} < x < 0 & + & \text{Upward} \\ \hline\\ \displaystyle 0 < x < \sqrt{\frac{3}{5}} & - & \text{Downward} \\ \hline\\ \displaystyle x > \sqrt{\frac{3}{5}} & + & \text{Upward}\\ \hline \end{array}$

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.

d.) Using the values obtained, illustrate the graph of $f$.