Anshid

Module IV — Graphing and Asymptotes

Module IV — Graphing with Derivatives and Asymptotes

Course: MAT1CJ101 — Differential Calculus
Hours: 10 | Textbook Sections: §3.4, §3.5 (Thomas & Finney, 9th Ed.)

1. Concavity and the Second Derivative

1.1 Concavity

The concavity of a graph describes whether it bends upward or downward.

  • Concave up on an interval: the graph bends upward (like a bowl). Tangent lines lie below the curve. $f’$ is increasing.
  • Concave down on an interval: the graph bends downward (like an upturned bowl). Tangent lines lie above the curve. $f’$ is decreasing.

Test using the second derivative:

\(f''(x) > 0 \text{ on } (a,b) \implies f \text{ is concave up on } (a,b)\) \(f''(x) < 0 \text{ on } (a,b) \implies f \text{ is concave down on } (a,b)\)

Why: $f’’ = (f’)’$. If $f’’ > 0$, then $f’$ is increasing — the slope increases, so the curve bends upward.

1.2 Inflection Points

An inflection point is a point where concavity changes (from up to down, or down to up).

Necessary condition: $f’‘(c) = 0$ or $f’‘(c)$ undefined at an inflection point $c$.

Important: $f’‘(c) = 0$ does NOT guarantee an inflection point. Check that concavity actually changes.

Worked Example 1: Find the inflection points of $f(x) = x^4 - 4x^3$.

\(f'(x) = 4x^3 - 12x^2\) \(f''(x) = 12x^2 - 24x = 12x(x - 2)\)

\[f''(x) = 0 \implies x = 0 \text{ or } x = 2\]

Sign of $f’’$:

  • $x < 0$: $12(-)(-) = + \implies$ concave up
  • $0 < x < 2$: $12(+)(-) = - \implies$ concave down
  • $x > 2$: $12(+)(+) = + \implies$ concave up

Concavity changes at $x = 0$ and $x = 2 \implies$ both are inflection points. $f(0) = 0$, $f(2) = 16 - 32 = -16$.

Inflection points: $(0, 0)$ and $(2, -16)$.

1.3 Graphing with $y’$ and $y’’$

To sketch a function, combine:

Information Source
Increasing/decreasing Sign of $f’$
Local maxima and minima Sign change of $f’$
Concave up/down Sign of $f’’$
Inflection points Sign change of $f’’$
Intercepts Set $y = 0$ (x-intercepts) and $x = 0$ (y-intercept)
Domain Identify any restrictions

Worked Example 2: Sketch $f(x) = x^3 - 3x$.

\(f'(x) = 3x^2 - 3 = 3(x-1)(x+1)\) \(f''(x) = 6x\)

$f’ = 0$ at $x = \pm 1$.

  • $f’ > 0$ for $x < -1$ and $x > 1 \implies$ increasing
  • $f’ < 0$ for $-1 < x < 1 \implies$ decreasing
  • $\implies$ local max at $x = -1$: $f(-1) = -1+3 = 2$
  • $\implies$ local min at $x = 1$: $f(1) = 1-3 = -2$

$f’’ = 0$ at $x = 0$.

  • $f’’ < 0$ for $x < 0 \implies$ concave down
  • $f’’ > 0$ for $x > 0 \implies$ concave up
  • $\implies$ inflection point at $(0, 0)$.

Intercepts: $f(0) = 0$; x-intercepts: $0 = x^3-3x = x(x^2-3)$, so $x = 0, \pm\sqrt{3}$.

Summary: rises from $-\infty$, local max at $(-1, 2)$, concave change at origin, local min at $(1, -2)$, rises to $+\infty$.

Worked Example 3: Graphing $y’$ and $y’’$ together — $f(x) = 6x - x^2 - x^3$.

\(f'(x) = 6 - 2x - 3x^2\) \(f''(x) = -2 - 6x = -2(1+3x)\)

$f’’ = 0 \implies x = -1/3$

  • $f’’ > 0$ for $x < -1/3 \implies$ concave up
  • $f’’ < 0$ for $x > -1/3 \implies$ concave down
  • Inflection point at $x = -1/3$: $f(-1/3) = -2 - 1/3 - (-1/3)^3 = -7/3 + 1/27 \approx -2.30$

For $f’$: $6 - 2x - 3x^2 = 0 \implies 3x^2 + 2x - 6 = 0$

\[x = \frac{-2 \pm \sqrt{4+72}}{6} = \frac{-2 \pm \sqrt{76}}{6}\]

2. The Second Derivative Test for Local Extrema

Theorem (Second Derivative Test): Suppose $f’(c) = 0$ and $f’‘(c)$ exists.

\(f''(c) > 0 \implies \text{local minimum at } c\) \(f''(c) < 0 \implies \text{local maximum at } c\) \(f''(c) = 0 \implies \text{test is inconclusive (use First Derivative Test instead)}\)

Intuition: If $f’(c) = 0$ (flat tangent) and $f’‘(c) > 0$ (concave up), the function bends upward from the flat point — so $c$ is a bottom (local min). Concave down $\implies$ top (local max).

Worked Example 4: Classify critical points of $f(x) = x^4 - 8x^2$.

\[f'(x) = 4x^3 - 16x = 4x(x^2-4), \text{ critical points at } x = 0, \pm 2\] \[f''(x) = 12x^2 - 16\]
  • $f’‘(0) = -16 < 0 \implies$ local maximum at $x = 0$, $f(0) = 0$
  • $f’‘(2) = 48-16 = 32 > 0 \implies$ local minimum at $x = 2$, $f(2) = 16-32 = -16$
  • $f’’(-2) = 48-16 = 32 > 0 \implies$ local minimum at $x = -2$, $f(-2) = -16$

Worked Example 5: Identify local extrema of $f(x) = x^3$ (fails the second derivative test).

$f’(x) = 3x^2$, $f’(0) = 0$. Critical point: $x = 0$. $f’‘(x) = 6x$, $f’‘(0) = 0$. Test inconclusive.

Use First Derivative Test: $f’(x) = 3x^2 \geq 0$ on both sides of 0 (no sign change). $\implies x = 0$ is NOT a local extremum (inflection point instead).

3. Limits at Infinity and Dominant Terms

3.1 Behaviour for Large x

The dominant term of a polynomial or rational function controls its behaviour as $ x \to \infty$.

For polynomials: The highest-degree term dominates.

\[f(x) = 5x^3 - 2x^2 + 7 \implies \text{behaves like } 5x^3 \text{ as } |x| \to \infty\]

For rational functions: The ratio of leading terms dominates.

Worked Example 6: Analyse the large-x behaviour of $f(x) = \dfrac{3x^2 - 5}{x^2 + 1}$.

Leading term: $3x^2/x^2 = 3$.

Divide top and bottom by $x^2$: \(\frac{3 - 5/x^2}{1 + 1/x^2}\)

As $x \to \pm\infty$: $\to 3/1 = 3$.

\[\lim_{x \to \pm\infty} f(x) = 3\]

Summary table for rational functions:

Let $f(x) = p(x)/q(x)$ where $\deg(p) = m$ and $\deg(q) = n$.

Comparison Limit at $\pm\infty$
$m < n$ 0
$m = n$ ratio of leading coefficients
$m > n$ $\pm\infty$ (or oscillates)

Worked Example 7: Find $\displaystyle\lim_{x \to \infty} \frac{x^2 + 3x}{x^3 - 1}$.

$\deg(\text{numerator}) = 2 < \deg(\text{denominator}) = 3$. Limit = 0.

3.2 End Behaviour

The end behaviour of $f$ describes what $f(x)$ does as $x \to +\infty$ and $x \to -\infty$.

Worked Example 8: End behaviour of $f(x) = 2x^3 - x$.

Leading term: $2x^3$.

  • As $x \to +\infty$: $2x^3 \to +\infty$, so $f(x) \to +\infty$.
  • As $x \to -\infty$: $2x^3 \to -\infty$, so $f(x) \to -\infty$.

4. Asymptotes

4.1 Horizontal Asymptotes

The line $y = L$ is a horizontal asymptote if:

\[\lim_{x \to +\infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L\]

Worked Example 9: Find horizontal asymptotes of $f(x) = \dfrac{2x + 1}{x - 3}$.

\[\lim_{x \to +\infty} \frac{2x+1}{x-3} = \lim_{x \to +\infty} \frac{2 + 1/x}{1 - 3/x} = 2/1 = 2\] \[\lim_{x \to -\infty} \frac{2x+1}{x-3} = 2 \quad \text{(same calculation)}\]

Horizontal asymptote: $y = 2$

Worked Example 10 (Sandwich Theorem for asymptotes): Find asymptotes of $y = 2 + \dfrac{\sin x}{x}$. (Oct 2024, Q17)

Since $-1 \leq \sin x \leq 1$ for all $x$: \(-\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \quad \text{for } x > 0\)

By the Squeeze Theorem: $\displaystyle\lim_{x \to +\infty} \frac{\sin x}{x} = 0$

Therefore: $\displaystyle\lim_{x \to +\infty} \left[2 + \frac{\sin x}{x}\right] = 2 + 0 = 2$

Similarly for $x \to -\infty$.

Horizontal asymptote: $y = 2$.

Note: $y = 2 + \dfrac{\sin x}{x}$ oscillates near $y = 2$ but approaches it at $\pm\infty$.

4.2 Vertical Asymptotes

The line $x = a$ is a vertical asymptote if:

\[\lim_{x \to a^+} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^-} f(x) = \pm\infty\]

For rational functions $f(x) = p(x)/q(x)$ in lowest terms: vertical asymptotes occur at zeros of $q(x)$.

Worked Example 11: Find vertical asymptotes of $f(x) = \dfrac{x+1}{x^2 - 4}$.

Denominator: $x^2 - 4 = (x-2)(x+2)$ Zeros at $x = 2$ and $x = -2$.

Numerator: $x+1 \neq 0$ at $x = 2$ or $x = -2$.

Vertical asymptotes: $x = 2$ and $x = -2$.

  • At $x = 2^+$: numerator $\to 3$ (positive), denominator $\to 0^+ \implies f \to +\infty$
  • At $x = 2^-$: numerator $\to 3$ (positive), denominator $\to 0^- \implies f \to -\infty$
  • At $x = -2^+$: numerator $\to -1$ (negative), denominator $\to 0^- \implies f \to +\infty$
  • At $x = -2^-$: numerator $\to -1$ (negative), denominator $\to 0^+ \implies f \to -\infty$

4.3 Oblique (Slant) Asymptotes

If $\deg(\text{numerator}) = \deg(\text{denominator}) + 1$, the rational function has an oblique asymptote — a slanted line $y = mx + b$ that the graph approaches at $\pm\infty$.

Find it by polynomial long division: $f(x) = (mx + b) + r(x)/q(x)$, where $r(x)/q(x) \to 0$.

Worked Example 12: Find the asymptote of $y = \dfrac{x^2 + 1}{x} = x + \dfrac{1}{x}$.

\[y = x + \frac{1}{x}\]

As $x \to \pm\infty$: $1/x \to 0$, so $y \to x$.

Oblique asymptote: $y = x$.

Vertical asymptote: $x = 0$.

5. Graphing with Asymptotes and Dominant Terms

5.1 Complete Curve Sketching Strategy

  1. Domain: identify restrictions, holes.
  2. Intercepts: x-intercepts (set $y=0$), y-intercept (set $x=0$).
  3. Symmetry: even? (y-axis), odd? (origin), neither?
  4. Asymptotes: horizontal (limits at $\pm\infty$), vertical (denominator = 0).
  5. Monotonicity: sign of $f’ \to$ increasing/decreasing intervals.
  6. Local extrema: First or Second Derivative Test.
  7. Concavity: sign of $f’’ \to$ concave up/down, inflection points.
  8. Sketch: plot key points, draw consistent with all information.

Worked Example 13: Graph $y = \dfrac{x^3 + 1}{x}$. (Oct 2024, Q20)

Step 1. Domain: $x \neq 0$. Rewrite: $y = x^2 + 1/x$.

Step 2. x-intercepts: $x^3 + 1 = 0 \implies x = -1$. y-intercept: none ($x = 0$ excluded).

Step 3. Symmetry: $f(-x) = -x^2 - 1/x = -(x^2 + 1/x) = -f(x)$. Odd function — symmetric about origin.

Step 4. Asymptotes:

  • Vertical: $x = 0$.
  • Oblique: $y = x^2 + 1/x \to x^2$ as $x \to \infty$ (not a line — it’s a parabola!).
  • More precisely: $f(x) - x^2 = 1/x \to 0$ as $x \to \infty$.
  • So $y = x^2$ is a “parabolic asymptote.”

Step 5. $y’ = 2x - 1/x^2 = (2x^3 - 1)/x^2$

  • $y’ = 0$: $2x^3 = 1 \implies x = (1/2)^{1/3} = 2^{-1/3} \approx 0.794$.
  • $y’$ undefined: $x = 0$.

Sign of $y’$:

  • $x < 0$: numerator $2x^3 - 1 < 0$ (since $x^3 < 0$), so $y’ < 0 \implies$ decreasing.
  • $0 < x < 2^{-1/3}$: $2x^3 < 1$, numerator $< 0 \implies y’ < 0 \implies$ decreasing.
  • $x > 2^{-1/3}$: numerator $> 0 \implies y’ > 0 \implies$ increasing.

Step 6. Local minimum at $x = 2^{-1/3}$: \(y = (2^{-1/3})^2 + \frac{1}{2^{-1/3}} = 2^{-2/3} + 2^{1/3} = 2^{-2/3}(1 + 2) = 3 \cdot 2^{-2/3} \approx 1.89\)

Step 7. $y’’ = 2 + 2/x^3 = 2(x^3+1)/x^3$

  • $y’’ = 0$: $x^3 = -1 \implies x = -1$.
  • $x < -1$: $x^3 < -1$, so $x^3+1 < 0$, numerator $< 0$, $x^3 < 0 \implies y’’ = 2(\text{neg})/(\text{neg}) > 0 \implies$ concave up.
  • $-1 < x < 0$: $x^3+1 > 0$ (since $x^3 > -1$), $x^3 < 0 \implies y’’ < 0 \implies$ concave down.
  • $x > 0$: $x^3+1 > 0$, $x^3 > 0 \implies y’’ > 0 \implies$ concave up.
  • Inflection at $x = -1$: $y(-1) = (-1+1)/(-1) = 0$. Point $(-1, 0)$ — the x-intercept.

Worked Example 14: Graph $y = 2 + \dfrac{\sin x}{x}$ and identify asymptotic behaviour.

From §3 above: horizontal asymptote $y = 2$ at both ends.

As $x \to 0$: $\sin(x)/x \to 1$ (well-known limit), so $y \to 3$ (no vertical asymptote).

For large $x$: $y$ oscillates between $2 - 1/ x $ and $2 + 1/ x $, converging to $y = 2$.

6. Constructing Functions from Limiting Behaviour

Worked Example 15: Find a function satisfying: (Oct 2024, Q18)

  • $\displaystyle\lim_{x \to \infty} f(x) = 1$
  • $\displaystyle\lim_{x \to -\infty} f(x) = -1$
  • $\displaystyle\lim_{x \to 1^+} f(x) = 1$
Horizontal asymptote $y = 1$ as $x \to +\infty$ and $y = -1$ as $x \to -\infty$ suggests $f(x) = x/ x $ scaled.

One natural candidate: $f(x) = \dfrac{x}{\sqrt{1+x^2}}$.

Check:

  • as $x \to +\infty$: $x/ x = x/x = 1$. ✓
  • as $x \to -\infty$: $x/ x = x/(-x) = -1$. ✓

At $x = 1$: $f(1) = 1/\sqrt{2} \approx 0.707$. If we also need $\lim_{x \to 1^+} f(x) = 1$, we need a function with a jump at $x = 1$.

Modified example with jump: \(f(x) = \begin{cases} x/\sqrt{1+x^2}, & x < 1 \\ 1, & x \geq 1 \end{cases}\)

Then: $\lim_{x \to 1^-} f(x) = 1/\sqrt{2}$, $\lim_{x \to 1^+} f(x) = 1$. (Jump discontinuity at $x = 1$.) The asymptotic behaviour is preserved.

Note: many answers are possible. The key is to show why limits govern the choice.

7. Summary

Concept Key Formula/Condition
Concave up $f’‘(x) > 0$ on interval
Concave down $f’‘(x) < 0$ on interval
Inflection point $f’’$ changes sign at $c$
Second Derivative Test $f’(c)=0$, $f’‘(c)>0 \implies$ local min; $f’‘(c)<0 \implies$ local max
Horizontal asymptote $\lim_{x \to \pm\infty} f(x) = L$, so $y = L$
Vertical asymptote $\lim_{x \to a} f(x) = \pm\infty$, so $x = a$
Oblique asymptote Long division when $\deg(\text{num}) = \deg(\text{den})+1$
Dominant term Highest-degree term governs large-x behaviour
Sandwich at $\infty$ Squeeze $(\sin x)/x$ between $\pm 1/x \to 0$

8. Practice Problems

Sec A style (3 marks each):

  1. Find the inflection points of $f(x) = x^3 - 6x$.

  2. Find the horizontal asymptotes of $f(x) = (3x - 1)/(x + 2)$.

  3. Find the vertical asymptotes of $f(x) = 1/(x^2 - 9)$. Determine the sign of $f$ on each side.

  4. Show that $\lim_{x \to \infty} 1/x = 0$. (Oct 2024, Q10)

  5. Use the Second Derivative Test to classify the critical points of $f(x) = x^3 - 3x$.

Sec B style (6 marks each):

  1. Using the Sandwich Theorem, find the horizontal asymptotes of $y = 2 + (\sin x)/x$. (Oct 2024, Q17)

  2. Find a function satisfying: $\lim_{x \to \infty} f(x) = 1$, $\lim_{x \to -\infty} f(x) = -1$, $\lim_{x \to 1^+} f(x) = 1$. Sketch the graph. (Oct 2024, Q18)

  3. For $f(x) = x^2/(x^2 - 4)$: (a) Find domain, intercepts, and asymptotes. (b) Find increasing/decreasing intervals and local extrema. (c) Find concavity and inflection points. Sketch the graph.

  4. Find all asymptotes (horizontal, vertical, oblique) of $f(x) = (x^2 + x - 1)/(x - 1)$.

Sec C style (10 marks):

  1. Graph $y = (x^3 + 1)/x$. Show all working: domain, intercepts, symmetry, asymptotes, monotonicity, local extrema, concavity, inflection points. (Oct 2024, Q20)