Lesson 24 (10/25/26)
Today: Graphing Functions
Office Hours: MWF 1:30pm – 2:30pm.
Meeting with Faculty: Tomorrow at 12pm, Windsor.
Lesson 24 (10/25/26)
Office Hours: MWF 1:30pm – 2:30pm.
Meeting with Faculty: Tomorrow at 12pm, Windsor.
First Derivative:
Note: \( f'(c) = 0 \) or DNE \( \implies x=c \) is a critical point.
Note: \( x=c \) is a local max/min if \( f' \) changes sign at \( x=c \).
Second Derivative:
Note: \( x=c \) is an inflection point if \( f''(c) = 0 \) & \( f'' \) changes sign at \( x=c \).
Example: \( f(x) = \frac{x^3}{3} - 16x \)
Find intervals where the function is increasing and decreasing.
\( f'(x) = x^2 - 16 \)
Critical points: \( x^2 - 16 = 0 \implies x = 4, x = -4 \)
Intervals of \( \uparrow \) & \( \downarrow \)
Increasing on \( (-\infty, -4), (4, \infty) \)
Decreasing on \( (-4, 4) \)
Sketch \( \frac{x^3}{3} - 16x \) only using the information from \( f' \)
\[ f(x) = \frac{x^3}{3} - 16x \]
Find intervals where the function is concave up and down.
\[ f'(x) = x^2 - 16 \]
\[ f''(x) = 2x \]
\[ f''(x) = 0 \implies x = 0 \]
Concave up on \( (0, \infty) \)
Concave down on \( (-\infty, 0) \)
\( x = 0 \) is inflection point
The upper portion of the diagram displays a horizontal number line with tick marks at \(-4\) and \(4\). Above the line, a series of vertical arrows indicate the direction of the slope of a function: upward-pointing arrows (increasing slope) appear to the left of \(-4\), downward-pointing arrows (decreasing slope) appear between \(-4\) and the midpoint, followed by a sequence of upward and downward arrows, and finally upward-pointing arrows to the right of \(4\). Below this line, there are two distinct wavy blue lines drawn: one segment is labeled "C. down" (concave down) spanning a region roughly between \(-4\) and the midpoint, and another segment is labeled "C. up" (concave up) spanning the region to the right of the midpoint.
More information about \( f(x) = \frac{x^3}{3} - 16x \)
Domain: \( (-\infty, \infty) \)
Intercepts:
X-intercept: \( \frac{x^3}{3} - 16x = 0 \)
\( \frac{x}{3} [x^2 - 48] = 0 \implies x = 0, x = \sqrt{48} = 4\sqrt{3} \)
\( x = -\sqrt{48} = -4\sqrt{3} \)
Y-intercept: \( x = 0 \implies \frac{x^3}{3} - 16x = 0 \)
No Vertical Asymptotes:
\( \lim_{x \to \pm\infty} f(x) = \pm\infty \implies \) No Horizontal Asymptotes
\( f(4) = \frac{64}{3} - 64 = -64 \times \frac{2}{3} \), \( f(-4) = \frac{2}{3} \times 64 \)
The following diagram provides a graphical representation of the cubic function \( f(x) = \frac{x^3}{3} - 16x \).
Domain: everything except \( x=1 \) & \( x=-1 \)
\( (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \)
Vertical Asymptotes/Holes: \( \lim_{x \to 1} f(x) \) & \( \lim_{x \to -1} f(x) \)
\[ \lim_{x \to 1^+} \frac{x^3}{x^2 - 1} = \frac{1}{\text{+ve, small}} = +\infty \]
\[ \lim_{x \to 1^-} \frac{x^3}{x^2 - 1} = \frac{1}{\text{-ve, small}} = -\infty \]
\[ \lim_{x \to -1^+} \frac{x^3}{x^2 - 1} = \frac{-1}{\text{-ve small}} = +\infty \]
\[ \lim_{x \to -1^-} \frac{x^3}{x^2 - 1} = \frac{-1}{\text{+ve small}} = -\infty \]
\[ \lim_{x \to \pm \infty} \frac{x^3}{x^2-1} \]
If \( f(x) = \frac{p(x)}{q(x)} \), if degree of \( p(x) = \) degree of \( q(x) + 1 \), \( f(x) \) has a slant asymptote.
Do Long Division to find Slant Asymptote
\[ \frac{x^3}{x^2-1} = \frac{x}{x^2-1} + x \]
Note: The term \( \frac{x}{x^2-1} \) has a limit of 0 as \( x \to \infty \).
\( y=x \) is a slant asymptote
\[ f(x) = \frac{x^3}{x^2 - 1} \]
x-intercept: \( f(x) = 0 \Rightarrow \frac{x^3}{x^2 - 1} = 0 \Rightarrow x = 0 \)
y-intercept: \( y = f(0) = \frac{0^3}{0^2 - 1} = 0 \)
Conclusion: The graph passes through \((0, 0)\) and does not intersect axes anywhere else.
Use \( f'(x) \) to find intervals of \( \uparrow \) and \( \downarrow \).
\[ f(x) = \frac{x^3}{x^2-1} \]
\[ f'(x) = \frac{(3x^2)(x^2-1) - x^3(2x)}{(x^2-1)^2} = \frac{3x^4 - 3x^2 - 2x^4}{(x^2-1)^2} = \frac{x^4 - 3x^2}{(x^2-1)^2} \]
Note: \( (x^2-1)^2 > 0 \) for \( x \neq \pm 1 \)
Numerator \( = x^4 - 3x^2 = x^2(x^2-3) = x^2(x-\sqrt{3})(x+\sqrt{3}) \)
\( f(x) = \dfrac{x^4 - 3x^2}{(x^2 - 1)^2} \)
\( f'(x) = \dfrac{(4x^3 - 6x)(x^2 - 1)^2 - (x^4 - 3x^2) \cdot (2(x^2 - 1) \cdot 2x)}{((x^2 - 1)^2)^2} \)
Note: Denominator \( > 0 \)
Numerator:
\( (x^2 - 1) [(4x^3 - 6x)(x^2 - 1) - (x^4 - 3x^2)(4x)] \)
\( (x^2 - 1) [4x^5 - 6x^3 - 4x^3 + 6x - 4x^5 + 12x^3] \)
\( (x^2 - 1)(2x^3 + 6x) \)
\( x(x^2 - 1)(2x^2 + 6) = x(x - 1)(x + 1) \cdot (2x^2 + 6) \)
Margin Note: \( \underbrace{2x^2 + 6}_{\ge 0} \)