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PAGE 1

4.4 Graphing Functions (part 1)

Sketching Guidelines

Domain
Symmetry → not very useful, ok to skip
Increasing / Decreasing Intervals
Relative max/min
deal with \( f' \)
Concave up/down intervals
Inflection points
deal with \( f'' \)
Asymptotes
Intercepts

finally, use the above info to sketch a graph

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Example

\( f(x) = x^3 - 12x^2 + 36x \)

Domain: \( (-\infty, \infty) \) because \( f(x) \) is a polynomial

Symmetry: skip

Inc / dec intervals:

\[ f'(x) = 3x^2 - 24x + 36 \]\[ = 3(x^2 - 8x + 12) \]\[ = 3(x - 6)(x - 2) \]

\( f' = 0 \rightarrow x = 6, x = 2 \)

\( f' \text{ DNE} \rightarrow \text{never, since } f' \text{ is polynomial} \)

Figure: A sign chart for f' showing a horizontal axis with points x=2 and x=6. Signs are + then - then +.

inc: \( (-\infty, 2), (6, \infty) \)

dec: \( (2, 6) \)

Relative max/min:

rel. max at \( x = 2, y = f(2) = 32 \rightarrow (2, 32) \)

rel. min at \( x = 6, y = f(6) = 0 \rightarrow (6, 0) \)

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Curve Sketching Analysis

Concave Up/Down Intervals

\[ f''(x) = 6x - 24 \]\[ f''(x) = 0 \rightarrow x = 4 \]\[ f'' \text{ DNE} \rightarrow \text{never} \]

The sign of \( f'' \) determines the concavity of the function. A sign chart shows the transition at \( x = 4 \).

CU (Concave Up): \( (4, \infty) \)
CD (Concave Down): \( (-\infty, 4) \)

Figure: Sign chart for f''(x) with a critical point at x=4, showing negative sign/concave down left and positive sign/concave up right.

Inflection Points and Asymptotes

Inflection pts: at \( x = 4 \), \( y = f(4) = 16 \rightarrow (4, 16) \)

Asymptotes: none, \( f(x) \) is a polynomial

Intercepts

x-intercept @ \( y = 0 \)

\[ f(x) = x^3 - 12x^2 + 36x = 0 \]\[ 0 = x(x^2 - 12x + 36) \]\[ 0 = x(x - 6)(x - 6) \]\[ x = 0, x = 6 \]

y-intercept @ \( x = 0 \)

\[ y = 0 \]

Summary of Known Points

x-ints: \( x = 0, x = 6 \)
y-int: \( y = 0 \)
rel. max: \( (2, 32) \)

rel. min: \( (6, 0) \)
inf. pt: \( (4, 16) \)

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Graphing the Function

Graph: Start with known points then use concavity or inc/dec info to fill in in between.

Figure: Graph of f(x) showing a local maximum at (2,32), inflection point at (4,16), and local minimum at (6,0).

Concavity Observations

Between \( x = 4 \) and \( x = 6 \): concave up
Beyond \( x = 6 \), also concave up
Left of \( x = 4 \), concave down

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Example: Function Analysis

\[ f(x) = \frac{x}{x^2 - 16} \]

Domain: \( x \neq 4, x \neq -4 \) (these are vertical asymptotes: denom = 0, numer \( \neq 0 \))

First Derivative Analysis

\[ f'(x) = \dots = -\frac{x^2 + 16}{(x^2 - 16)^2} \]

\( f' = 0 \rightarrow x^2 + 16 = 0 \rightarrow \text{no solutions} \)
\( f' \text{ DNE} \rightarrow x^2 - 16 = 0 \rightarrow x = -4, x = 4 \)

Sign of \( f' \):

Figure: Sign chart for f' showing negative signs and downward arrows across intervals separated by x = -4 and x = 4.

dec: \( (-\infty, -4), (-4, 4), (4, \infty) \)
inc: none

Rel max / min: none, since there are no \( f' \) sign changes

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CU/CD Intervals (Concavity)

\[ f''(x) = \dots = \frac{2x(x^2 + 48)}{(x^2 - 16)^3} \]

\( f'' = 0 \rightarrow 2x(x^2 + 48) = 0 \rightarrow x = 0 \) (\( x^2 + 48 = 0 \) has no solutions)
\( f'' \text{ DNE} \rightarrow x^2 - 16 = 0 \rightarrow x = -4, x = 4 \)

Sign of \( f'' \):

Figure: Sign chart for f'' with intervals: negative (CD), positive (CU), negative (CD), positive (CU) split at -4, 0, 4.

CU: \( (-4, 0), (4, \infty) \)
CD: \( (-\infty, -4), (0, 4) \)

Inflection Points

At \( x = 0 \) only because even though \( f'' \) changes sign at \( x = -4, x = 4 \), \( x = \pm 4 \) are not in domain so are NOT points on the graph.

inf. pt: \( (0, 0) \) \( \leftarrow f(0) \)

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Asymptotes and Intercepts Analysis

Asymptotes

Vertical: \( x = 4, x = -4 \)

Horizontal: \( \lim_{x \to \infty} f(x) \) and \( \lim_{x \to -\infty} f(x) \)

\( \lim_{x \to \infty} \frac{x}{x^2 - 16} = 0 \)

\( \lim_{x \to -\infty} \frac{x}{x^2 - 16} = 0 \)

\( y = 0 \) horiz. asyn. on both sides

Intercepts

x-ints @ \( y = 0 \)

\( f(x) = \frac{x}{x^2 - 16} \)

\( 0 = \frac{x}{x^2 - 16} \rightarrow x = 0 \)

y-ints @ \( x = 0 \)

\( f(x) = \frac{x}{x^2 - 16} \)

\( y = \frac{0}{-16} \rightarrow y = 0 \)

Points we know:

\( x = 0 \) (x-int)
\( y = 0 \) (y-int)
\( (0,0) \) (inf. pt)

Same point

Now we use asymptotes to help us

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Graphing the Function

Use CU/CD (concave up/concave down) to fill in between the asymptotes.

Graph of f(x) = \frac{x}{x^2-16} with vertical asymptotes at x=-4 and x=4 , and an inflection point at (0,0) . — Figure: Graph of \( f(x) = \frac{x}{x^2-16} \) with vertical asymptotes at \( x=-4 \) and \( x=4 \), and an inflection point at \( (0,0) \).

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Example: Function Analysis

Analyze the function:

\[ f(x) = x - \sin x \quad \text{on} \quad [-2\pi, 2\pi] \]

Domain: \( [-2\pi, 2\pi] \)

Increasing / Decreasing Intervals

\[ f'(x) = 1 - \cos x \]\[ f' = 0 \rightarrow \cos x = 1 \rightarrow x = -2\pi, x = 0, x = 2\pi \]

The sign chart for \( f' \) shows that the derivative is non-negative across the domain, with zeros at the critical points.

Figure: Sign chart for f' from -2π to 2π showing positive intervals and increasing arrows.

Rel. max/min: none (no \( f' \) sign change)

Concavity (CU/CD)

\[ f''(x) = \sin x \]\[ f'' = 0 \rightarrow \sin x = 0 \rightarrow x = -2\pi, -\pi, 0, \pi, 2\pi \]

The sign chart for \( f'' \) indicates alternating concavity between the roots of the second derivative.

Figure: Sign chart for f'' from -2π to 2π showing alternating concavity symbols (up and down).

Inflection points: \( x = -\pi, x = 0, x = \pi \) which correspond to the points:

\[ (-\pi, -\pi), (0, 0), (\pi, \pi) \]

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Asymptotes: none

Intercepts

x-int (\( y=0 \)) \( \rightarrow x = 0 \)
y-int (\( x=0 \)) \( \rightarrow y = 0 \)

Summary of Known Points

Inflection points: \( (-\pi, -\pi), (0, 0), (\pi, \pi) \)
Intercepts: \( (0, 0) \)
End points:
- \( x = -2\pi, y = -2\pi \rightarrow (-2\pi, -2\pi) \)
- \( x = 2\pi, y = 2\pi \rightarrow (2\pi, 2\pi) \)

Graph of \( f(x) = x - \sin x \)

The graph shows a monotonically increasing function with inflection points at multiples of \( \pi \).

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Some functions have slant asymptotes

\[ f(x) = \frac{x^2 + 15}{5x + 1} \]

Notice as \( x \to \infty \), \( x^2 + 15 \approx x^2 \), and \( 5x + 1 \approx 5x \).

So,

\[ \frac{x^2 + 15}{5x + 1} \to \frac{x^2}{5x} \to \frac{1}{5}x \]

This means the graph will approach \( y = \frac{1}{5}x \) as \( x \to \pm \infty \).

Vertical Asymptote

vertical: \( x = -\frac{1}{5} \)

Concavity Analysis

Go through CU/CD step:

CU: \( (-\frac{1}{5}, \infty) \)
CD: \( (-\infty, -\frac{1}{5}) \)

Figure: Coordinate graph showing a function with a vertical asymptote at x = -1/5 and a slant asymptote y = 1/5x.