The Hook: Reading the Story of a Graph
Three friends go on a road trip through Spain. Their journey has:
- Uphill stretches (exciting, growing)
- Downhill slopes (relaxing, declining)
- Sharp turns (sudden changes in direction)
- Smooth curves (gradual transitions)
- Plateaus (stable periods)
Just like their journey, every function $f(x)$ tells a story through its graph!
Curve sketching is the art of reading this story using calculus:
- Where is $f$ increasing/decreasing? (Uphill/downhill)
- Where does $f$ change direction? (Critical points)
- Where does $f$ curve up/down? (Concavity)
- Where does $f$ approach infinity? (Asymptotes)
Master curve sketching, and you can visualize any function instantly!
Interactive: Build a Curve Step-by-Step
See how derivatives control the shape of curves!
The Curve Sketching Checklist
Complete Curve Sketching Procedure:
Step 1: Domain & Range — Where is $f$ defined? Any restrictions?
Step 2: Intercepts
- $y$-intercept: Set $x = 0$
- $x$-intercept: Solve $f(x) = 0$
Step 3: Symmetry
- Even function: $f(-x) = f(x)$ (symmetric about $y$-axis)
- Odd function: $f(-x) = -f(x)$ (symmetric about origin)
Step 4: Asymptotes
- Vertical: Where denominator = 0
- Horizontal: $\lim_{x \to \pm\infty} f(x)$
- Oblique: Long division if degree(numerator) = degree(denominator) + 1
Step 5: First Derivative — Monotonicity
- Find $f'(x)$ and critical points ($f'(x) = 0$ or undefined)
- Sign chart for $f'(x)$ → Intervals of increase/decrease
Step 6: Second Derivative — Concavity
- Find $f''(x)$ and inflection point candidates ($f''(x) = 0$ or undefined)
- Sign chart for $f''(x)$ → Concave up/down regions
Step 7: Key Points — Evaluate $f$ at:
- Critical points
- Inflection points
- A few sample points
Step 8: Sketch — Combine all information!
Monotonicity (Increasing/Decreasing)
Definitions
Increasing on an interval:
- Strictly increasing: $f(x_1) < f(x_2)$ whenever $x_1 < x_2$
- Non-decreasing: $f(x_1) \leq f(x_2)$ whenever $x_1 < x_2$
Decreasing on an interval:
- Strictly decreasing: $f(x_1) > f(x_2)$ whenever $x_1 < x_2$
- Non-increasing: $f(x_1) \geq f(x_2)$ whenever $x_1 < x_2$
Using First Derivative
Monotonicity Test:
On interval $I$:
- $f'(x) > 0$ for all $x \in I$ $\Rightarrow$ $f$ is strictly increasing on $I$
- $f'(x) < 0$ for all $x \in I$ $\Rightarrow$ $f$ is strictly decreasing on $I$
- $f'(x) = 0$ for all $x \in I$ $\Rightarrow$ $f$ is constant on $I$
Visual:
- Positive slope → Graph rises from left to right
- Negative slope → Graph falls from left to right
- Zero slope → Horizontal line
Example: Finding Monotonicity Intervals
Find where $f(x) = x^3 - 6x^2 + 9x + 1$ is increasing/decreasing.
Step 1: Find $f'(x)$
$$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$$Step 2: Critical points
$$f'(x) = 0 \Rightarrow x = 1, 3$$Step 3: Sign chart
| Interval | $x < 1$ | $x = 1$ | $1 < x < 3$ | $x = 3$ | $x > 3$ |
|---|---|---|---|---|---|
| $(x-1)$ | $-$ | $0$ | $+$ | $+$ | $+$ |
| $(x-3)$ | $-$ | $-$ | $-$ | $0$ | $+$ |
| $f'(x)$ | $+$ | $0$ | $-$ | $0$ | $+$ |
| $f(x)$ | ↗ | Max | ↘ | Min | ↗ |
Answer:
- Increasing: $(-\infty, 1) \cup (3, \infty)$
- Decreasing: $(1, 3)$
- Local maximum at $x = 1$
- Local minimum at $x = 3$
Concavity and Inflection Points
Definitions
Concave Up (∪): Graph curves upward, like a cup
- Tangent lines lie below the curve
- Function “holds water”
Concave Down (∩): Graph curves downward, like a cap
- Tangent lines lie above the curve
- Function “spills water”
Inflection Point: Point where concavity changes (from ∪ to ∩ or vice versa)
Using Second Derivative
Concavity Test:
On interval $I$:
- $f''(x) > 0$ for all $x \in I$ $\Rightarrow$ $f$ is concave up on $I$
- $f''(x) < 0$ for all $x \in I$ $\Rightarrow$ $f$ is concave down on $I$
Inflection Point at $x = c$:
- $f''(c) = 0$ (or undefined)
- $f''$ changes sign at $x = c$
Note: $f''(c) = 0$ is necessary but NOT sufficient! Must check sign change.
Example: Finding Concavity and Inflection Points
Find concavity and inflection points of $f(x) = x^4 - 4x^3$
Step 1: Find $f''(x)$
$$f'(x) = 4x^3 - 12x^2$$ $$f''(x) = 12x^2 - 24x = 12x(x - 2)$$Step 2: Solve $f''(x) = 0$
$$x = 0, 2$$Step 3: Sign chart
| Interval | $x < 0$ | $x = 0$ | $0 < x < 2$ | $x = 2$ | $x > 2$ |
|---|---|---|---|---|---|
| $12x$ | $-$ | $0$ | $+$ | $+$ | $+$ |
| $(x-2)$ | $-$ | $-$ | $-$ | $0$ | $+$ |
| $f''(x)$ | $+$ | $0$ | $-$ | $0$ | $+$ |
| Concavity | ∪ | ? | ∩ | ? | ∪ |
Step 4: Check sign changes
- At $x = 0$: $f''$ changes from $+$ to $-$ → Inflection point ✓
- At $x = 2$: $f''$ changes from $-$ to $+$ → Inflection point ✓
Values:
- $f(0) = 0$
- $f(2) = 16 - 32 = -16$
Answer:
- Concave up: $(-\infty, 0) \cup (2, \infty)$
- Concave down: $(0, 2)$
- Inflection points: $(0, 0)$ and $(2, -16)$
Asymptotes
Vertical Asymptotes
Vertical asymptote at $x = a$:
If $\lim_{x \to a^+} f(x) = \pm\infty$ or $\lim_{x \to a^-} f(x) = \pm\infty$
Finding: Solve denominator = 0 (for rational functions)
Example: $f(x) = \frac{1}{x - 2}$
Denominator = 0: $x = 2$
$\lim_{x \to 2^+} \frac{1}{x-2} = +\infty$, $\lim_{x \to 2^-} \frac{1}{x-2} = -\infty$
Vertical asymptote: $x = 2$
Horizontal Asymptotes
Horizontal asymptote $y = L$:
If $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$
For rational functions $\frac{p(x)}{q(x)}$:
- Degree of $p <$ Degree of $q$ → $y = 0$
- Degree of $p =$ Degree of $q$ → $y = \frac{\text{leading coeff of } p}{\text{leading coeff of } q}$
- Degree of $p >$ Degree of $q$ → No horizontal asymptote
Examples:
$f(x) = \frac{2x^2 + 1}{x^2 - 4}$: Same degree → $y = \frac{2}{1} = 2$
$f(x) = \frac{3x}{x^2 + 1}$: Numerator degree < denominator → $y = 0$
$f(x) = \frac{x^3}{x^2 - 1}$: Numerator degree > denominator → No horizontal asymptote (but has oblique)
Oblique (Slant) Asymptotes
Oblique asymptote $y = mx + c$:
When degree(numerator) = degree(denominator) + 1
Finding: Perform polynomial long division
Example: $f(x) = \frac{x^2 + 1}{x}$
Long division: $\frac{x^2 + 1}{x} = x + \frac{1}{x}$
As $x \to \infty$, $\frac{1}{x} \to 0$
Oblique asymptote: $y = x$
Complete Example: Full Curve Sketch
Sketch $f(x) = \frac{x^2}{x^2 - 4}$
Step 1: Domain
Denominator ≠ 0: $x^2 - 4 \neq 0 \Rightarrow x \neq \pm 2$
Domain: $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$
Step 2: Intercepts
- $y$-intercept: $f(0) = 0$ → $(0, 0)$
- $x$-intercept: $x^2 = 0 \Rightarrow x = 0$ → $(0, 0)$
Step 3: Symmetry
$$f(-x) = \frac{(-x)^2}{(-x)^2 - 4} = \frac{x^2}{x^2 - 4} = f(x)$$Even function → Symmetric about $y$-axis
Step 4: Asymptotes
Vertical: $x = -2, 2$
Horizontal:
$$\lim_{x \to \infty} \frac{x^2}{x^2 - 4} = \lim_{x \to \infty} \frac{1}{1 - 4/x^2} = 1$$Horizontal asymptote: $y = 1$
Step 5: First Derivative (Monotonicity)
$$f'(x) = \frac{2x(x^2 - 4) - x^2(2x)}{(x^2 - 4)^2} = \frac{2x^3 - 8x - 2x^3}{(x^2-4)^2} = \frac{-8x}{(x^2-4)^2}$$Critical point: $x = 0$
Sign chart:
| Interval | $x < -2$ | $-2 < x < 0$ | $0 < x < 2$ | $x > 2$ |
|---|---|---|---|---|
| $-8x$ | $+$ | $+$ | $-$ | $-$ |
| $(x^2-4)^2$ | $+$ | $+$ | $+$ | $+$ |
| $f'(x)$ | $+$ | $+$ | $-$ | $-$ |
| $f(x)$ | ↗ | ↗ | ↘ | ↘ |
Increasing: $(-\infty, -2) \cup (-2, 0)$ Decreasing: $(0, 2) \cup (2, \infty)$ Local maximum at $x = 0$: $f(0) = 0$
Step 6: Second Derivative (Concavity)
$$f''(x) = \frac{-8(x^2-4)^2 - (-8x) \cdot 2(x^2-4)(2x)}{(x^2-4)^4}$$ $$= \frac{-8(x^2-4) + 32x^2}{(x^2-4)^3} = \frac{24x^2 + 32}{(x^2-4)^3}$$Numerator $24x^2 + 32 > 0$ always!
Sign depends on $(x^2 - 4)^3$:
| Interval | $x < -2$ | $-2 < x < 2$ | $x > 2$ |
|---|---|---|---|
| $(x^2-4)^3$ | $-$ | $+$ (wait, $x^2 < 4$ so $x^2-4 < 0$!) | Actually $-$ when $-2 < x < 2$, $+$ otherwise |
Correct analysis:
For $x \in (-2, 2)$: $x^2 - 4 < 0$, so $(x^2-4)^3 < 0$ → $f''(x) < 0$ → Concave down
For $x < -2$ or $x > 2$: $x^2 - 4 > 0$, so $(x^2-4)^3 > 0$ → $f''(x) > 0$ → Concave up
Step 7: Key Points
- Origin: $(0, 0)$ — intercept and local max
- Behavior near asymptotes:
- As $x \to 2^+$: $f(x) \to +\infty$
- As $x \to 2^-$: $f(x) \to -\infty$
- As $x \to -2$: Similar (by symmetry)
Step 8: Sketch Summary
- Symmetric about $y$-axis (even function)
- Passes through origin (local max)
- Vertical asymptotes at $x = \pm 2$
- Horizontal asymptote at $y = 1$
- Increasing to the left of $x = 0$, decreasing to the right
- Concave up for $|x| > 2$, concave down for $|x| < 2$
Special Curve Types
Polynomial Curves
- Degree $n$: At most $n - 1$ turning points
- Even degree: Same end behavior (both up or both down)
- Odd degree: Opposite end behavior (one up, one down)
Rational Functions
- Check for vertical asymptotes (denominator = 0)
- Check for horizontal/oblique asymptotes (end behavior)
- Often have symmetry
Trigonometric Curves
- Periodic (repeat every $2\pi$ for sin/cos)
- Bounded between -1 and 1 (for standard trig)
- Use derivatives to find max/min within one period
Exponential/Logarithmic
- $e^x$: Always increasing, concave up, horizontal asymptote $y = 0$ as $x \to -\infty$
- $\ln x$: Always increasing, concave down, vertical asymptote $x = 0$
Memory Tricks
- $f''(x) > 0$: ∪ (Concave UP like a CUP — holds water)
- $f''(x) < 0$: ∩ (Concave down like a CAP — spills water)
“Positive second derivative = Cup!”
$f'(x)$ tells the slope story:
- $f' > 0$: Walking uphill (increasing)
- $f' < 0$: Walking downhill (decreasing)
- $f' = 0$: Flat ground (critical point — could be peak, valley, or plateau)
“Follow the slope!”
Denominator = 0 → Vertical Degrees same → Horizontal (ratio of leading coefficients) Degree of top = degree of bottom + 1 → Oblique (long division)
“Three D’s: Denominator, Degrees, Degree+1”
$f''(c) = 0$ is not enough for inflection point!
Must check: Does $f''$ change sign at $x = c$?
Example: $f(x) = x^4$ at $x = 0$: $f''(0) = 0$ but NO sign change (always concave up) → NOT an inflection point!
Practice Problems
Level 1: Foundation (NCERT)
Question: Find intervals where $f(x) = x^2 - 4x + 3$ is increasing/decreasing.
Solution:
$$f'(x) = 2x - 4 = 0 \Rightarrow x = 2$$- For $x < 2$: $f'(x) < 0$ → Decreasing
- For $x > 2$: $f'(x) > 0$ → Increasing
Answer: Decreasing on $(-\infty, 2)$, increasing on $(2, \infty)$
Question: Find concavity of $f(x) = x^3 - 3x$
Solution:
$$f'(x) = 3x^2 - 3$$ $$f''(x) = 6x$$- $f''(x) > 0$ when $x > 0$ → Concave up on $(0, \infty)$
- $f''(x) < 0$ when $x < 0$ → Concave down on $(-\infty, 0)$
Inflection point: $(0, 0)$ (sign change in $f''$)
Level 2: JEE Main
Question: Find asymptotes of $f(x) = \frac{2x^2 - 3}{x^2 - 1}$
Solution:
Vertical: $x^2 - 1 = 0 \Rightarrow x = \pm 1$
Horizontal: Same degree → $y = \frac{2}{1} = 2$
Answer: Vertical asymptotes $x = -1, 1$; horizontal asymptote $y = 2$
Question: Find inflection points of $f(x) = x^4 - 6x^2$
Solution:
$$f'(x) = 4x^3 - 12x$$ $$f''(x) = 12x^2 - 12 = 12(x^2 - 1) = 12(x-1)(x+1)$$$f''(x) = 0$ at $x = \pm 1$
Sign chart:
| Interval | $x < -1$ | $-1 < x < 1$ | $x > 1$ |
|---|---|---|---|
| $f''(x)$ | $+$ | $-$ | $+$ |
Both $x = -1$ and $x = 1$ have sign changes!
Inflection points: $(-1, f(-1)) = (-1, -5)$ and $(1, -5)$
Level 3: JEE Advanced
Question: Sketch the complete curve of $f(x) = \frac{x}{x^2 + 1}$
Solution:
Domain: All real numbers (denominator never zero)
Intercepts: $f(0) = 0$ → $(0, 0)$
Symmetry: $f(-x) = \frac{-x}{x^2+1} = -f(x)$ → Odd function (origin symmetry)
Asymptotes:
- Vertical: None
- Horizontal: Degree of numerator < denominator → $y = 0$
First derivative:
$$f'(x) = \frac{(x^2+1) - x(2x)}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}$$$f'(x) = 0$ at $x = \pm 1$
- $x < -1$: $f' < 0$ (decreasing)
- $-1 < x < 1$: $f' > 0$ (increasing)
- $x > 1$: $f' < 0$ (decreasing)
Local min at $x = -1$: $f(-1) = -\frac{1}{2}$ Local max at $x = 1$: $f(1) = \frac{1}{2}$
Second derivative:
$$f''(x) = \frac{-2x(x^2+1)^2 - (1-x^2) \cdot 2(x^2+1)(2x)}{(x^2+1)^4}$$ $$= \frac{-2x(x^2+1) - (1-x^2)(4x)}{(x^2+1)^3} = \frac{-2x^3 - 2x - 4x + 4x^3}{(x^2+1)^3}$$ $$= \frac{2x^3 - 6x}{(x^2+1)^3} = \frac{2x(x^2 - 3)}{(x^2+1)^3}$$$f''(x) = 0$ at $x = 0, \pm\sqrt{3}$
Inflection points at these three values.
Sketch:
- Passes through origin
- Symmetric about origin (odd)
- Max at $(1, 1/2)$, min at $(-1, -1/2)$
- Horizontal asymptote $y = 0$
- Three inflection points
Question: Find the range of values of $a$ for which $f(x) = x^3 - 3ax + 2$ is strictly increasing for all $x \in \mathbb{R}$.
Solution:
For strictly increasing: $f'(x) > 0$ for all $x$
$$f'(x) = 3x^2 - 3a$$For $f'(x) > 0$ for all $x$:
$$3x^2 - 3a > 0 \Rightarrow x^2 > a$$Since minimum value of $x^2$ is $0$:
For $x^2 > a$ to hold for all $x$, we need $a < 0$ (actually $a \leq 0$)
Wait, let’s reconsider: $f'(x) = 3x^2 - 3a \geq 0$ for all $x$
Since $x^2 \geq 0$, minimum of $3x^2$ is $0$ (at $x = 0$)
So: $0 - 3a \geq 0 \Rightarrow a \leq 0$
But for strictly increasing (not just non-decreasing), we need $f'(x) > 0$ except possibly at isolated points.
If $a = 0$: $f'(x) = 3x^2 \geq 0$ (equals 0 only at $x = 0$) → Still strictly increasing overall ✓
If $a < 0$: $f'(x) = 3x^2 - 3a > 0$ always (since $x^2 \geq 0$ and $-3a > 0$) ✓
If $a > 0$: $f'(x) = 0$ at $x = \pm\sqrt{a}$, and $f' < 0$ between these roots → NOT always increasing ✗
Answer: $a \leq 0$
Common Mistakes to Avoid
$f''(c) = 0$ is necessary but not sufficient!
Must check: Does $f''$ change sign at $x = c$?
Counter-example: $f(x) = x^4$ at $x = 0$: $f''(0) = 0$ but concave up everywhere → NOT an inflection point!
When finding asymptotes or sketching, always check the domain first!
Can’t have a curve where the function isn’t defined.
Don’t assume symmetry without checking!
Test $f(-x)$:
- If $f(-x) = f(x)$: Even (y-axis symmetry)
- If $f(-x) = -f(x)$: Odd (origin symmetry)
- Otherwise: No symmetry
Oblique asymptotes only exist when degree(num) = degree(den) + 1
If degrees differ by more than 1, or numerator degree is smaller, no oblique asymptote!
Quick Revision Box
| Concept | Test | Result |
|---|---|---|
| Increasing | $f'(x) > 0$ | Function rises |
| Decreasing | $f'(x) < 0$ | Function falls |
| Concave up | $f''(x) > 0$ | ∪ (cup shape) |
| Concave down | $f''(x) < 0$ | ∩ (cap shape) |
| Inflection point | $f''(x) = 0$ AND sign change | Concavity switches |
| Vertical asymptote | Denominator = 0 | $x = a$ |
| Horizontal asymptote | $\lim_{x \to \infty} f(x)$ | $y = L$ |
| Oblique asymptote | Deg(num) = Deg(den) + 1 | $y = mx + c$ |
JEE Strategy Tips
Weightage: Curve sketching concepts appear in 3-4 JEE questions (combined with other topics).
Time-Saver: Use symmetry! If $f$ is even/odd, only analyze half the graph, then reflect.
Common Pattern: JEE loves rational functions with multiple asymptotes and turning points.
Trap Alert: Always verify inflection points with sign change test, don’t just solve $f'' = 0$!
Advanced Tip: For complex curves, plot key points first (intercepts, max/min, inflection) then connect smoothly.
Exam Hack: If asked to sketch without calculator, focus on qualitative features (increasing/decreasing, concavity, asymptotes) rather than exact coordinates.
Teacher’s Summary
First derivative $f'$ controls slope and monotonicity: $f' > 0$ → increasing, $f' < 0$ → decreasing
Second derivative $f''$ controls concavity: $f'' > 0$ → concave up (∪), $f'' < 0$ → concave down (∩)
Inflection points require $f'' = 0$ (or undefined) AND sign change in $f''$
Asymptotes reveal end behavior:
- Vertical: Denominator = 0
- Horizontal: Compare degrees of num/den
- Oblique: Long division when deg(num) = deg(den) + 1
Curve sketching checklist: Domain → Intercepts → Symmetry → Asymptotes → $f'$ analysis → $f''$ analysis → Key points → Sketch!
“Every curve tells a story — derivatives help you read it!”
Related Topics
Within Limits, Continuity & Differentiability
- Differentiability — Foundation for using derivatives
- Higher Derivatives — Second derivatives for concavity
- Maxima-Minima — Critical points in curve sketching
- Mean Value Theorems — Monotonicity proofs
Applications
- Limits — Finding asymptotes
- Integration — Area under curves
Mathematical Foundations
- Functions — Understanding function behavior
- Coordinate Geometry — Graphing techniques
Physics Connections
- Kinematics — Position-time graphs
- Potential Energy Curves — Analyzing energy landscapes