Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 8

Use the guidelines of curve sketching to sketch the curve. $y =(4-x^2)^5$

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 $y$-intercpt, when $x = 0$
$y = \left(4 - (0)^2 \right)^5 = 1024$
Solving for $x$-intercept, when $y = 0$
$0 = (4-x^2)^5$
We have, $x = \pm 2$

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

D. Asymptotes.
None, the function has no denominator

E. Intervals of Increase or Decrease.
If we take the derivative of $f(x)$. By using Chain Rule,

$
\begin{equation}
\begin{aligned}
f'(x) &= 5(4-x^2)^4(-2x)\\
\\
f'(x) &= -10x(4-x^2)^4
\end{aligned}
\end{equation}
$

When $f'(x) = 0$, $0 = -10x (4-x^2)^4$
We have, $x =0$ and $-x^2 + 4 = 0$
The critical numbers are, $x = 0$ and $x = \pm 2$
So, the intervals of increase or decrease are...

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f\\
\hline\\
x < -2 & + & \text{increasing on } (-\infty,-2)\\
\hline\\
-2 < x < 0 & + & \text{increasing on } (-2,0)\\
\hline\\
0 < x < 2& - & \text{decreasing 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 $x=0$, $f(0) = 1024$ is a local maximum. However, since $f'(x)$ does not change from negative to positive, it means that the function has no local minimum.

G. Concavity and Points of Inflection.

$
\begin{equation}
\begin{aligned}
\text{if } f'(x) &= -10x(4-x^2)^4, \text{ then, by using Product Rule and Chain Rule }\\
\\
f''(x) &= -10 x \cdot 4(4-x^2)^3 (-2x) + (-10) (4-x^2)^4\\
\\
f''(x) &= -10(4-x^2)^3 \left[ -8x^2 + (4-x^2) \right]\\
\\
f''(x) &= -10(4-x^2)^3 \left[ -9x^2 + 4 \right]
\end{aligned}
\end{equation}
$

When $f''(x) =0$
$0 = -10(4-x^2)^3 \left[ -9x^2 + 4 \right]$
We have, $(4-x^2)^3 = 0$ and $-9x^2 + 4 = 0$
The inflection poitns are, $x = \pm 2$ and $\displaystyle x = \pm \frac{2}{3}$

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 < -2 & - & \text{Downward}\\
\hline\\
-2 < x < \frac{-2}{3} & + & \text{Upward}\\
\hline\\
\frac{-2}{3} < x < \frac{2}{3} & - & \text{Downward}\\
\hline\\
\frac{2}{3} < x < 2 & + & \text{Upward}\\
\hline\\
x > 2 & - & \text{Downward}\\
\hline
\end{array}
$


H. Sketch the Graph.