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

Use the guidelines of curve sketching to sketch the curve. $y =8x^2 - x^4$

The guidelines of Curve Sketching

A. Domain.
We know that $f(x)$ is a polynomial having a domain of $(-\infty, \infty)$

B. Intercepts.
Solving for $y$-intercept, when $x = 0$.
$y = 8 (0)^2 - (0)^4 = 0$
Solving for $x$-intercept, when $y=0$,

$\begin{equation} \begin{aligned} 0 &= 8x^2 - x^4\\ \\ 0 &= x^2(8-x^2) \end{aligned} \end{equation}$


we have, $x = 0$ and $x^2 - 8 =0$
So the $x$-intercept are $x = 0$, $x = 2.8284$ and $x = -2.8284$


C. Symmetry.
$f(-x) = f(x)$, therefore the function is symmetric to $y$-axis.

D. Asymptotes.
None.

E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$, we have $f'(x) = 16x - 4x^3$

$\begin{equation} \begin{aligned} \text{when } f'(x) &= 0\\ \\ 0 &= 16x-4x^3\\ \\ 0 &= 4x(4-x^2)\\ \\ \\ \\ \text{we have, }\\ \\ x &= 0 \text{ and } x^2 - 4 = 0\\ \\ \text{The critical numbers are, } \\ \\ x &= 0 \text{ and } x = \pm 2 \end{aligned} \end{equation}$

So, the intervals of increase or decrease are...

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



F. Local Maximum and Minimum Values.
Since $f'(x)$ changes from positive to negative at $x = 2$ and $x = -2$, then $f(2) = 16$ and $f(-2) = 16$ are local maximum. On the other hand, since $f'(x)$ changes from negative to positive at $x = 0$, then $f(0) = 0$ is local minimum.


G. Concavity and Points of Inflection.

$\begin{equation} \begin{aligned} \text{if } f'(x) &= 16x - 4x^3, \text{ then }\\ \\ f''(x) &= 16 - 12x^2\\ \\ \\ \\ \text{when } f''(x) &= 0,\\ \\ 0 & = 16 - 12x^2 \end{aligned} \end{equation}$

Thus, the concavity can be determined by dividing the interval to...

$\begin{array}{|c|c|c|} \hline\\ \text{Interval} & f''(x) & \text{Concavity}\\ \hline\\ x < -1.1547 & - & \text{Downward}\\ \hline\\ -1.1547 < x < 1.1547 & + & \text{Upward}\\ \hline\\ x > 1.1547 & - & \text{Downward}\\ \hline \end{array}$


H. Sketch the Graph