Curve Sketching

Using derivatives to sketch accurate graphs of functions

📈 Curve Sketching

What is Curve Sketching?

Curve sketching is the art of drawing an accurate graph of a function using calculus. Instead of plotting points, we use derivatives to understand the function's behavior!

💡 Key Idea: Use f(x)f'(x) and f(x)f''(x) to determine where the function is increasing/decreasing and where it's curving up/down.


The Complete Curve Sketching Checklist

Step-by-Step Guide

Step 1: Find the domain

Step 2: Find intercepts

  • yy-intercept: Set x=0x = 0
  • xx-intercepts: Solve f(x)=0f(x) = 0

Step 3: Check for symmetry

  • Even: f(x)=f(x)f(-x) = f(x) (symmetric about yy-axis)
  • Odd: f(x)=f(x)f(-x) = -f(x) (symmetric about origin)

Step 4: Find asymptotes

  • Vertical: Where denominator = 0 (for rational functions)
  • Horizontal: Check limx±f(x)\lim_{x \to \pm\infty} f(x)
  • Slant: If degree of numerator is 1 more than denominator

Step 5: Find critical points (f(x)=0f'(x) = 0 or undefined)

Step 6: Determine intervals of increase/decrease

  • f(x)>0f'(x) > 0: increasing
  • f(x)<0f'(x) < 0: decreasing

Step 7: Find local extrema (use First or Second Derivative Test)

Step 8: Find inflection points (f(x)=0f''(x) = 0 and concavity changes)

Step 9: Determine concavity

  • f(x)>0f''(x) > 0: concave up ∪
  • f(x)<0f''(x) < 0: concave down ∩

Step 10: Sketch the graph using all information!


What Each Derivative Tells You

First Derivative: f(x)f'(x)

Sign of f(x)f'(x):

  • f(x)>0f'(x) > 0 → function increasing
  • f(x)<0f'(x) < 0 → function decreasing
  • f(x)=0f'(x) = 0horizontal tangent (potential max/min)

Critical points: Where f(x)=0f'(x) = 0 or f(x)f'(x) undefined

Local extrema: Where ff' changes sign

Second Derivative: f(x)f''(x)

Sign of f(x)f''(x):

  • f(x)>0f''(x) > 0concave up ∪ (curving upward)
  • f(x)<0f''(x) < 0concave down ∩ (curving downward)
  • f(x)=0f''(x) = 0 → potential inflection point

Inflection points: Where concavity changes (and ff'' changes sign)


Example: Complete Curve Sketch

Sketch f(x)=x36x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1


Step 1: Domain

Polynomial → domain is all real numbers: (,)(-\infty, \infty)


Step 2: Intercepts

yy-intercept: f(0)=1f(0) = 1 → point (0,1)(0, 1)

xx-intercepts: x36x2+9x+1=0x^3 - 6x^2 + 9x + 1 = 0 (hard to solve, skip for now)


Step 3: Symmetry

f(x)=x36x29x+1f(x)f(-x) = -x^3 - 6x^2 - 9x + 1 \neq f(x) and f(x)\neq -f(x)

No symmetry


Step 4: Asymptotes

Polynomial → no asymptotes


Step 5: First derivative and critical points

f(x)=3x212x+9=3(x24x+3)=3(x1)(x3)f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

Critical points: x=1x = 1 and x=3x = 3


Step 6: Sign chart for f(x)f'(x)

        1         3
   ++++  |  ----  |  ++++
      ↗  ▼  ↘     ▲  ↗
  • Increasing on (,1)(-\infty, 1) and (3,)(3, \infty)
  • Decreasing on (1,3)(1, 3)

Step 7: Local extrema

At x=1x = 1: ff' changes + to − → local max f(1)=16+9+1=5f(1) = 1 - 6 + 9 + 1 = 5

At x=3x = 3: ff' changes − to + → local min f(3)=2754+27+1=1f(3) = 27 - 54 + 27 + 1 = 1


Step 8: Second derivative and inflection points

f(x)=6x12=6(x2)f''(x) = 6x - 12 = 6(x - 2)

f(x)=0f''(x) = 0 when x=2x = 2


Step 9: Concavity

           2
   ----    |    ++++
      ∩         ∪
  • Concave down on (,2)(-\infty, 2)
  • Concave up on (2,)(2, \infty)
  • Inflection point at x=2x = 2: f(2)=824+18+1=3f(2) = 8 - 24 + 18 + 1 = 3

Step 10: Sketch

Key points:

  • (0,1)(0, 1) - yy-intercept
  • (1,5)(1, 5) - local max
  • (2,3)(2, 3) - inflection point
  • (3,1)(3, 1) - local min

The graph:

  • Increases to (1,5)(1, 5), then decreases to (3,1)(3, 1), then increases again
  • Concave down until x=2x = 2, then concave up
  • Passes through (0,1)(0, 1)

Special Features to Look For

Cusps and Corners

Where ff is continuous but ff' doesn't exist

Example: f(x)=xf(x) = |x| has a corner at x=0x = 0

Vertical Tangents

Where f(x)=±f'(x) = \pm\infty

Example: f(x)=x1/3f(x) = x^{1/3} has vertical tangent at x=0x = 0

Discontinuities

  • Jump discontinuity
  • Removable discontinuity
  • Infinite discontinuity (vertical asymptote)

Analyzing Rational Functions

For f(x)=p(x)q(x)f(x) = \frac{p(x)}{q(x)}:

Vertical Asymptotes

Occur where q(x)=0q(x) = 0 (denominator zero, numerator non-zero)

Example: f(x)=1x2f(x) = \frac{1}{x-2} has vertical asymptote at x=2x = 2

Horizontal Asymptotes

Compare degrees of p(x)p(x) and q(x)q(x):

  1. If deg(pp) < deg(qq): y=0y = 0 is horizontal asymptote
  2. If deg(pp) = deg(qq): y=leading coef of pleading coef of qy = \frac{\text{leading coef of } p}{\text{leading coef of } q}
  3. If deg(pp) > deg(qq): No horizontal asymptote (may have slant asymptote)

Example: f(x)=2x2+1x24f(x) = \frac{2x^2 + 1}{x^2 - 4} has horizontal asymptote y=2y = 2

Slant (Oblique) Asymptotes

When deg(pp) = deg(qq) + 1, use polynomial division

Example: f(x)=x2+1xf(x) = \frac{x^2 + 1}{x} has slant asymptote y=xy = x


Quick Summary Table

| Feature | How to Find | What It Tells You | |---------|-------------|-------------------| | Domain | Where ff is defined | Valid xx values | | Intercepts | f(0)f(0) and f(x)=0f(x) = 0 | Where graph crosses axes | | Critical pts | f(x)=0f'(x) = 0 or undef | Potential max/min | | Inc/Dec | Sign of f(x)f'(x) | Direction of graph | | Local extrema | Sign change in ff' | Peaks and valleys | | Inflection pts | f(x)=0f''(x) = 0 + sign change | Where concavity changes | | Concavity | Sign of f(x)f''(x) | Curvature direction | | Asymptotes | Limits, denominators | Boundary behavior |


⚠️ Common Mistakes

Mistake 1: Not Checking Sign Changes

f(c)=0f''(c) = 0 doesn't guarantee an inflection point - concavity must change!

Mistake 2: Forgetting Domain

Always consider where the function is actually defined.

Mistake 3: Plotting Only Critical Points

Include inflection points, intercepts, and other key features!

Mistake 4: Wrong Asymptote Analysis

Vertical asymptotes: denominator = 0 Horizontal asymptotes: check limits at infinity

Mistake 5: Ignoring Behavior at Infinity

Always check what happens as x±x \to \pm\infty


The Big Picture

Putting It All Together

  1. f(x)f(x) tells you the height of the graph
  2. f(x)f'(x) tells you the slope (increasing/decreasing)
  3. f(x)f''(x) tells you the curvature (concave up/down)

All three work together to give you a complete picture!


Shortcut for Simple Polynomials

For polynomials, you can often skip some steps:

  • No asymptotes
  • Domain is always R\mathbb{R}
  • Continuous everywhere
  • Focus on critical points, extrema, and inflection points

📝 Practice Strategy

  1. Follow the checklist systematically
  2. Make a sign chart for both ff' and ff''
  3. Calculate key points (don't just mark xx-values, find yy too!)
  4. Sketch lightly first, then refine
  5. Check your work: Does the sketch match your analysis?
  6. Label everything: maxima, minima, inflection points, asymptotes

📚 Practice Problems

1Problem 1medium

Question:

Sketch the graph of f(x)=x44x3f(x) = x^4 - 4x^3 using calculus.

💡 Show Solution

Step 1: Domain

Polynomial → (,)(-\infty, \infty)


Step 2: Intercepts

yy-intercept: f(0)=0f(0) = 0(0,0)(0, 0)

xx-intercepts: x44x3=0x^4 - 4x^3 = 0 x3(x4)=0x^3(x - 4) = 0 x=0x = 0 or x=4x = 4(0,0)(0, 0) and (4,0)(4, 0)


Step 3: Symmetry

f(x)=x4+4x3f(x)f(-x) = x^4 + 4x^3 \neq f(x) → No symmetry


Step 4: First derivative

f(x)=4x312x2=4x2(x3)f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)

Critical points: x=0,3x = 0, 3


Step 5: Sign chart for ff'

Test points:

  • x=1x = -1: f(1)=4(1)(4)=16<0f'(-1) = 4(1)(-4) = -16 < 0
  • x=1x = 1: f(1)=4(1)(2)=8<0f'(1) = 4(1)(-2) = -8 < 0
  • x=4x = 4: f(4)=4(16)(1)=64>0f'(4) = 4(16)(1) = 64 > 0
        0         3
   ----  |  ----  |  ++++
      ↘       ↘  ▲  ↗

Decreasing on (,3)(-\infty, 3), increasing on (3,)(3, \infty)

Local minimum at x=3x = 3: f(3)=81108=27f(3) = 81 - 108 = -27

Note: x=0x = 0 is NOT an extremum (no sign change)


Step 6: Second derivative

f(x)=12x224x=12x(x2)f''(x) = 12x^2 - 24x = 12x(x - 2)

f(x)=0f''(x) = 0 when x=0x = 0 or x=2x = 2


Step 7: Concavity

Test points:

  • x=1x = -1: f(1)=12(1)(3)=36...wait,letmerecalculatef''(-1) = 12(1)(-3) = -36... wait, let me recalculate
  • x=1x = -1: f(1)=12(1)(12)=12(1)(3)=36>0f''(-1) = 12(-1)(-1-2) = 12(-1)(-3) = 36 > 0
  • x=1x = 1: f(1)=12(1)(12)=12(1)(1)=12<0f''(1) = 12(1)(1-2) = 12(1)(-1) = -12 < 0
  • x=3x = 3: f(3)=12(3)(32)=36>0f''(3) = 12(3)(3-2) = 36 > 0
        0         2
   ++++  |  ----  |  ++++
      ∪       ∩       ∪

Inflection points at x=0x = 0 and x=2x = 2

f(0)=0f(0) = 0, f(2)=1632=16f(2) = 16 - 32 = -16


Summary for sketch:

Key points:

  • (0,0)(0, 0) - intercept and inflection point
  • (2,16)(2, -16) - inflection point
  • (3,27)(3, -27) - local minimum
  • (4,0)(4, 0) - xx-intercept

Behavior:

  • Decreasing from -\infty to (3,27)(3, -27)
  • Then increasing to ++\infty
  • Concave up, then down, then up again
  • Passes through origin

2Problem 2hard

Question:

Sketch the graph of g(x)=x2x24g(x) = \frac{x^2}{x^2 - 4} using calculus. Include all asymptotes.

💡 Show Solution

Step 1: Domain

Denominator: x24=(x2)(x+2)=0x^2 - 4 = (x-2)(x+2) = 0 when x=±2x = \pm 2

Domain: (,2)(2,2)(2,)(-\infty, -2) \cup (-2, 2) \cup (2, \infty)


Step 2: Intercepts

yy-intercept: g(0)=04=0g(0) = \frac{0}{-4} = 0(0,0)(0, 0)

xx-intercepts: x2=0x^2 = 0x=0x = 0(0,0)(0, 0)


Step 3: Symmetry

g(x)=(x)2(x)24=x2x24=g(x)g(-x) = \frac{(-x)^2}{(-x)^2 - 4} = \frac{x^2}{x^2-4} = g(x)

EVEN function - symmetric about yy-axis!


Step 4: Asymptotes

Vertical: At x=2x = -2 and x=2x = 2 (where denominator = 0)

Horizontal: limxx2x24=limx114/x2=1\lim_{x \to \infty} \frac{x^2}{x^2 - 4} = \lim_{x \to \infty} \frac{1}{1 - 4/x^2} = 1

Horizontal asymptote: y=1y = 1


Step 5: First derivative (Quotient Rule)

g(x)=2x(x24)x2(2x)(x24)2g'(x) = \frac{2x(x^2-4) - x^2(2x)}{(x^2-4)^2}

=2x38x2x3(x24)2= \frac{2x^3 - 8x - 2x^3}{(x^2-4)^2}

=8x(x24)2= \frac{-8x}{(x^2-4)^2}

Critical point: 8x=0-8x = 0x=0x = 0


Step 6: Sign of gg'

Since (x24)2>0(x^2-4)^2 > 0 always, sign depends on 8x-8x:

  • x<0x < 0: 8x>0-8x > 0 → increasing
  • x>0x > 0: 8x<0-8x < 0 → decreasing

Local maximum at x=0x = 0: g(0)=0g(0) = 0


Step 7: Second derivative (skip for brevity)

Can verify concavity, but we have enough information.


Step 8: Behavior near asymptotes

Near x=2+x = 2^+: numerator → 4, denominator → 0+0^+g(x)+g(x) \to +\infty

Near x=2x = 2^-: numerator → 4, denominator → 00^-g(x)g(x) \to -\infty

By symmetry, similar behavior at x=2x = -2


Summary for sketch:

  • Symmetric about yy-axis
  • Vertical asymptotes at x=±2x = \pm 2
  • Horizontal asymptote at y=1y = 1
  • Local max at (0,0)(0, 0)
  • Increasing on (,2)(-\infty, -2) and (2,0)(-2, 0)
  • Decreasing on (0,2)(0, 2) and (2,)(2, \infty)
  • As x±x \to \pm\infty, g(x)1g(x) \to 1

3Problem 3expert

Question:

Analyze and sketch h(x)=xexh(x) = xe^{-x} completely.

💡 Show Solution

Step 1: Domain

All real numbers: (,)(-\infty, \infty)


Step 2: Intercepts

yy-intercept: h(0)=0e0=0h(0) = 0 \cdot e^0 = 0(0,0)(0, 0)

xx-intercepts: xex=0xe^{-x} = 0x=0x = 0(0,0)(0, 0)


Step 3: Limits (behavior at infinity)

As xx \to -\infty: exe^{-x} \to \infty, so h(x)h(x) \to -\infty

As x+x \to +\infty: xex0\frac{x}{e^x} \to 0 (exponential dominates)

Horizontal asymptote: y=0y = 0 (as xx \to \infty)


Step 4: First derivative (Product Rule)

h(x)=(1)(ex)+(x)(ex)h'(x) = (1)(e^{-x}) + (x)(-e^{-x})

=exxex= e^{-x} - xe^{-x}

=ex(1x)= e^{-x}(1 - x)

Critical point: 1x=01 - x = 0x=1x = 1


Step 5: Sign of hh'

ex>0e^{-x} > 0 always, so sign depends on (1x)(1-x):

  • x<1x < 1: (1x)>0(1-x) > 0 → increasing
  • x>1x > 1: (1x)<0(1-x) < 0 → decreasing

Local maximum at x=1x = 1: h(1)=1e1=1e0.368h(1) = 1 \cdot e^{-1} = \frac{1}{e} \approx 0.368


Step 6: Second derivative

h(x)=(ex)(1x)+ex(1)h''(x) = (-e^{-x})(1-x) + e^{-x}(-1)

=ex(1x)ex= -e^{-x}(1-x) - e^{-x}

=ex(1x+1)= -e^{-x}(1-x+1)

=ex(2x)= -e^{-x}(2-x)

=ex(x2)= e^{-x}(x-2)

Inflection point: x2=0x - 2 = 0x=2x = 2


Step 7: Concavity

  • x<2x < 2: (x2)<0(x-2) < 0 → concave down
  • x>2x > 2: (x2)>0(x-2) > 0 → concave up

Inflection point at x=2x = 2: h(2)=2e2=2e20.271h(2) = 2e^{-2} = \frac{2}{e^2} \approx 0.271


Summary for sketch:

Key points:

  • (0,0)(0, 0) - origin
  • (1,1/e)(1, 1/e) - local maximum
  • (2,2/e2)(2, 2/e^2) - inflection point

Behavior:

  • Increases from -\infty to max at x=1x=1
  • Decreases from max, approaching 0 as xx \to \infty
  • Concave down until x=2x=2, then concave up
  • Horizontal asymptote y=0y=0 on the right

Answer: The graph rises from negative infinity through the origin, reaches a maximum at (1,1/e)(1, 1/e), has an inflection point at (2,2/e2)(2, 2/e^2), and approaches 0 as xx \to \infty.