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