Curve Sketching

Use calculus to understand and draw function graphs

Before calculators and graphing software, mathematicians sketched curves by hand using logic and calculus. While we now have powerful technology at our fingertips, understanding how to analyze a function and predict its shape remains one of the most valuable skills in mathematics. When you sketch a curve, you are not just drawing a picture – you are telling the story of how a function behaves. Where does it rise? Where does it fall? Where does it bend? These questions connect calculus to the real world, where understanding the shape of a curve can mean predicting market trends, modeling population growth, or designing bridges that will not collapse.

The good news is that curve sketching is not about memorizing formulas – it is about building a systematic checklist that you can apply to any function. By the end of this lesson, you will have a reliable method for understanding any function’s behavior, even before you see its graph.

Core Concepts

Domain and Range

Before you can sketch a curve, you need to know where the function actually exists. The domain tells you which $x$-values are allowed as inputs, while the range tells you which $y$-values the function can produce.

Common domain restrictions include:

  • Division by zero: exclude values where the denominator equals zero
  • Square roots of negative numbers: for real-valued functions, the expression under the radical must be non-negative
  • Logarithms of non-positive numbers: $\ln(x)$ requires $x > 0$

For example, the function $f(x) = \frac{1}{x-2}$ has domain $(-\infty, 2) \cup (2, \infty)$ because we cannot divide by zero when $x = 2$.

Finding the range often requires more work – sometimes you need to analyze the function’s behavior first, then determine what $y$-values are achievable.

Intercepts

Intercepts are the points where your curve crosses the axes. They give you concrete anchor points to build your sketch around.

$y$-intercept: Set $x = 0$ and solve for $y$. There can be at most one $y$-intercept (if $x = 0$ is in the domain).

$x$-intercepts (zeros): Set $y = 0$ and solve for $x$. There can be multiple $x$-intercepts, or none at all.

For $f(x) = x^2 - 4$:

  • $y$-intercept: $f(0) = -4$, so the point is $(0, -4)$
  • $x$-intercepts: $x^2 - 4 = 0 \Rightarrow x = \pm 2$, so the points are $(-2, 0)$ and $(2, 0)$

Symmetry

Checking for symmetry can save you half your work. If a function has symmetry, you only need to sketch one portion and reflect it.

Even functions satisfy $f(-x) = f(x)$. Their graphs are symmetric about the $y$-axis. Examples include $f(x) = x^2$, $f(x) = \cos(x)$, and $f(x) = |x|$.

Odd functions satisfy $f(-x) = -f(x)$. Their graphs are symmetric about the origin – if you rotate 180 degrees, the graph looks the same. Examples include $f(x) = x^3$, $f(x) = \sin(x)$, and $f(x) = \frac{1}{x}$.

Most functions are neither even nor odd, but it is always worth checking.

Asymptotes

Asymptotes are lines that the curve approaches but never quite reaches. They describe the function’s behavior at extreme values or near points where it is undefined.

Vertical asymptotes occur at values where the function approaches infinity. For rational functions, look for values where the denominator equals zero (but the numerator does not).

For $f(x) = \frac{1}{x-3}$, there is a vertical asymptote at $x = 3$ because: $$\lim_{x \to 3^+} \frac{1}{x-3} = +\infty \quad \text{and} \quad \lim_{x \to 3^-} \frac{1}{x-3} = -\infty$$

Horizontal asymptotes describe end behavior as $x \to \pm\infty$. For rational functions $\frac{p(x)}{q(x)}$:

  • If degree of $p$ < degree of $q$: horizontal asymptote at $y = 0$
  • If degree of $p$ = degree of $q$: horizontal asymptote at $y = \frac{\text{leading coefficient of } p}{\text{leading coefficient of } q}$
  • If degree of $p$ > degree of $q$: no horizontal asymptote (but there may be a slant asymptote)

Slant (oblique) asymptotes occur when the numerator’s degree is exactly one more than the denominator’s. Perform polynomial long division to find the asymptote.

Intervals of Increase and Decrease

This is where calculus becomes essential. The first derivative tells you whether a function is going up or down.

  • When $f’(x) > 0$, the function is increasing (moving upward as you go right)
  • When $f’(x) < 0$, the function is decreasing (moving downward as you go right)
  • When $f’(x) = 0$, the function has a horizontal tangent (possible maximum, minimum, or inflection point)

To find intervals of increase and decrease:

  1. Compute $f’(x)$
  2. Find where $f’(x) = 0$ or is undefined – these are your critical points
  3. Test the sign of $f’(x)$ in each interval between critical points
  4. Conclude: positive means increasing, negative means decreasing

Local Extrema

Local maxima and local minima are the peaks and valleys of your function. They occur at critical points where the derivative equals zero or is undefined.

First Derivative Test: At a critical point $x = c$:

  • If $f’(x)$ changes from positive to negative, then $f(c)$ is a local maximum
  • If $f’(x)$ changes from negative to positive, then $f(c)$ is a local minimum
  • If $f’(x)$ does not change sign, then $f(c)$ is neither (it might be an inflection point)

Second Derivative Test: At a critical point $x = c$ where $f’(c) = 0$:

  • If $f’’(c) > 0$, then $f(c)$ is a local minimum (concave up means the critical point is a valley)
  • If $f’’(c) < 0$, then $f(c)$ is a local maximum (concave down means the critical point is a peak)
  • If $f’’(c) = 0$, the test is inconclusive – use the first derivative test instead

Concavity

Concavity describes how the curve bends. The second derivative controls this behavior.

  • When $f’’(x) > 0$, the function is concave up – it curves like a cup that holds water
  • When $f’’(x) < 0$, the function is concave down – it curves like an arch that spills water

Think of concave up as a smile and concave down as a frown. Or imagine driving on a road: concave up feels like the bottom of a valley, while concave down feels like the top of a hill.

To find intervals of concavity:

  1. Compute $f’’(x)$
  2. Find where $f’’(x) = 0$ or is undefined
  3. Test the sign of $f’’(x)$ in each interval
  4. Conclude: positive means concave up, negative means concave down

Inflection Points

An inflection point is where the concavity changes – the curve transitions from bending one way to bending the other way. At an inflection point, the function crosses its own tangent line.

To find inflection points:

  1. Find where $f’’(x) = 0$ or is undefined
  2. Check that $f’’(x)$ actually changes sign at that point
  3. Calculate the $y$-value: the inflection point is $(c, f(c))$

Important: Not every point where $f’’(x) = 0$ is an inflection point. The concavity must actually change. For example, $f(x) = x^4$ has $f’’(0) = 0$, but $f’’(x) = 12x^2 \geq 0$ everywhere, so there is no sign change and no inflection point at $x = 0$.

Putting It All Together: The Complete Checklist

When you approach a curve sketching problem, follow this systematic checklist:

  1. Domain: Where is the function defined?
  2. Intercepts: Find $x$-intercepts (set $y = 0$) and $y$-intercept (set $x = 0$)
  3. Symmetry: Check if $f(-x) = f(x)$ (even) or $f(-x) = -f(x)$ (odd)
  4. Asymptotes: Find vertical, horizontal, and slant asymptotes
  5. First derivative: Find $f’(x)$, critical points, and intervals of increase/decrease
  6. Local extrema: Classify critical points as maxima, minima, or neither
  7. Second derivative: Find $f’’(x)$ and intervals of concavity
  8. Inflection points: Where does concavity change?
  9. Sketch: Plot key points and connect with appropriate curves

Notation and Terminology

Term Meaning Example
Increasing $f’(x) > 0$ Slope going up
Decreasing $f’(x) < 0$ Slope going down
Concave up $f’’(x) > 0$ “Holds water”
Concave down $f’’(x) < 0$ “Spills water”
Inflection point Concavity changes $f’’(x) = 0$ and sign change
Critical point $f’(x) = 0$ or undefined Potential extremum
Local maximum Peak in the graph $f’$ changes from $+$ to $-$
Local minimum Valley in the graph $f’$ changes from $-$ to $+$
Vertical asymptote $\lim_{x \to a} f(x) = \pm\infty$ Graph approaches vertical line
Horizontal asymptote $\lim_{x \to \pm\infty} f(x) = L$ Graph levels off

Examples

Example 1: Intervals of Increase and Decrease

Problem: Determine where $f(x) = x^3$ is increasing and decreasing.

Solution:

Step 1: Find the first derivative. $$f’(x) = 3x^2$$

Step 2: Find critical points by setting $f’(x) = 0$. $$3x^2 = 0 \Rightarrow x = 0$$

Step 3: Analyze the sign of $f’(x)$.

Since $3x^2 \geq 0$ for all $x$, and $3x^2 > 0$ for all $x \neq 0$:

  • For $x < 0$: $f’(x) = 3x^2 > 0$ (increasing)
  • For $x > 0$: $f’(x) = 3x^2 > 0$ (increasing)

Conclusion: $f(x) = x^3$ is increasing on $(-\infty, 0)$ and $(0, \infty)$. In fact, it is increasing on all of $(-\infty, \infty)$.

Note that even though $f’(0) = 0$, this is not a local extremum – the derivative does not change sign. The point $(0, 0)$ is actually an inflection point.

Example 2: Concavity of a Simple Function

Problem: Find the intervals of concavity for $f(x) = x^2$ and identify any inflection points.

Solution:

Step 1: Find the first derivative. $$f’(x) = 2x$$

Step 2: Find the second derivative. $$f’’(x) = 2$$

Step 3: Analyze the sign of $f’’(x)$.

Since $f’’(x) = 2 > 0$ for all $x$, the function is concave up everywhere.

Step 4: Check for inflection points.

There are no points where $f’’(x) = 0$, so there are no inflection points.

Conclusion: $f(x) = x^2$ is concave up on $(-\infty, \infty)$ with no inflection points. This confirms what we know about parabolas opening upward – they form a U-shape that “holds water.”

Example 3: Complete Curve Sketch

Problem: Sketch the curve $f(x) = x^3 - 3x^2$.

Solution:

Step 1: Domain This polynomial is defined for all real numbers: $(-\infty, \infty)$.

Step 2: Intercepts

$y$-intercept: $f(0) = 0^3 - 3(0)^2 = 0$. Point: $(0, 0)$

$x$-intercepts: Set $f(x) = 0$: $$x^3 - 3x^2 = 0$$ $$x^2(x - 3) = 0$$ $$x = 0 \text{ or } x = 3$$ Points: $(0, 0)$ and $(3, 0)$

Step 3: Symmetry

Check: $f(-x) = (-x)^3 - 3(-x)^2 = -x^3 - 3x^2$

This is neither $f(x)$ nor $-f(x)$, so no symmetry.

Step 4: Asymptotes

Polynomials have no vertical or horizontal asymptotes. As $x \to \infty$, $f(x) \to \infty$. As $x \to -\infty$, $f(x) \to -\infty$.

Step 5: First Derivative Analysis

$$f’(x) = 3x^2 - 6x = 3x(x - 2)$$

Critical points: $f’(x) = 0$ when $x = 0$ or $x = 2$.

Sign analysis:

  • For $x < 0$: $f’(x) = 3(\text{neg})(\text{neg}) > 0$ (increasing)
  • For $0 < x < 2$: $f’(x) = 3(\text{pos})(\text{neg}) < 0$ (decreasing)
  • For $x > 2$: $f’(x) = 3(\text{pos})(\text{pos}) > 0$ (increasing)

Step 6: Local Extrema

At $x = 0$: $f’$ changes from $+$ to $-$, so local maximum at $(0, f(0)) = (0, 0)$.

At $x = 2$: $f’$ changes from $-$ to $+$, so local minimum at $(2, f(2)) = (2, -4)$.

Step 7: Second Derivative Analysis

$$f’’(x) = 6x - 6 = 6(x - 1)$$

$f’’(x) = 0$ when $x = 1$.

Sign analysis:

  • For $x < 1$: $f’’(x) < 0$ (concave down)
  • For $x > 1$: $f’’(x) > 0$ (concave up)

Step 8: Inflection Point

At $x = 1$, concavity changes from down to up. Inflection point: $(1, f(1)) = (1, 1 - 3) = (1, -2)$

Step 9: Sketch

Key points to plot:

  • Local max: $(0, 0)$
  • Local min: $(2, -4)$
  • Inflection: $(1, -2)$
  • $x$-intercept: $(3, 0)$

The curve starts from $-\infty$ (lower left), increases to the local max at the origin, decreases through the inflection point to the local min at $(2, -4)$, then increases toward $+\infty$.

Example 4: Rational Function with Asymptotes

Problem: Sketch the curve $f(x) = \frac{x}{x^2 + 1}$.

Solution:

Step 1: Domain

The denominator $x^2 + 1 > 0$ for all $x$, so the domain is $(-\infty, \infty)$.

Step 2: Intercepts

$y$-intercept: $f(0) = \frac{0}{1} = 0$. Point: $(0, 0)$

$x$-intercepts: $\frac{x}{x^2 + 1} = 0$ when $x = 0$. Point: $(0, 0)$

Step 3: Symmetry

$$f(-x) = \frac{-x}{(-x)^2 + 1} = \frac{-x}{x^2 + 1} = -f(x)$$

The function is odd, so it has origin symmetry.

Step 4: Asymptotes

No vertical asymptotes (denominator never zero).

Horizontal asymptote: As $x \to \pm\infty$: $$\lim_{x \to \pm\infty} \frac{x}{x^2 + 1} = \lim_{x \to \pm\infty} \frac{1/x}{1 + 1/x^2} = \frac{0}{1} = 0$$

Horizontal asymptote: $y = 0$

Step 5: First Derivative Analysis

Using the quotient rule: $$f’(x) = \frac{(1)(x^2 + 1) - x(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}$$

Critical points: $f’(x) = 0$ when $1 - x^2 = 0$, so $x = \pm 1$.

Sign analysis (note: denominator is always positive):

  • For $|x| < 1$: $1 - x^2 > 0$, so $f’(x) > 0$ (increasing)
  • For $|x| > 1$: $1 - x^2 < 0$, so $f’(x) < 0$ (decreasing)

Step 6: Local Extrema

At $x = -1$: $f’$ changes from $-$ to $+$, so local minimum. $f(-1) = \frac{-1}{2}$. Local min at $(-1, -\frac{1}{2})$.

At $x = 1$: $f’$ changes from $+$ to $-$, so local maximum. $f(1) = \frac{1}{2}$. Local max at $(1, \frac{1}{2})$.

Step 7: Second Derivative Analysis

$$f’(x) = \frac{1 - x^2}{(x^2 + 1)^2}$$

Using the quotient rule again: $$f’’(x) = \frac{-2x(x^2+1)^2 - (1-x^2) \cdot 2(x^2+1)(2x)}{(x^2+1)^4}$$

Simplifying: $$f’’(x) = \frac{-2x(x^2+1) - 4x(1-x^2)}{(x^2+1)^3} = \frac{-2x^3 - 2x - 4x + 4x^3}{(x^2+1)^3} = \frac{2x^3 - 6x}{(x^2+1)^3} = \frac{2x(x^2 - 3)}{(x^2+1)^3}$$

$f’’(x) = 0$ when $x = 0$ or $x = \pm\sqrt{3}$.

Step 8: Inflection Points

Testing signs confirms concavity changes at $x = -\sqrt{3}$, $x = 0$, and $x = \sqrt{3}$.

Inflection points at:

  • $(-\sqrt{3}, f(-\sqrt{3})) = (-\sqrt{3}, -\frac{\sqrt{3}}{4})$
  • $(0, 0)$
  • $(\sqrt{3}, \frac{\sqrt{3}}{4})$

Step 9: Sketch

The curve passes through the origin with odd symmetry, rises to a maximum at $(1, \frac{1}{2})$, then approaches the horizontal asymptote $y = 0$ from above. By symmetry, it has a minimum at $(-1, -\frac{1}{2})$ and approaches $y = 0$ from below as $x \to -\infty$.

Example 5: Comprehensive Analysis of a Quartic

Problem: Sketch $f(x) = x^4 - 4x^3$ with all features labeled.

Solution:

Step 1: Domain

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

Step 2: Intercepts

$y$-intercept: $f(0) = 0$. Point: $(0, 0)$

$x$-intercepts: $$x^4 - 4x^3 = 0$$ $$x^3(x - 4) = 0$$ $$x = 0 \text{ (multiplicity 3) or } x = 4$$ Points: $(0, 0)$ and $(4, 0)$

Note: The multiplicity of 3 at $x = 0$ means the curve will have an inflection-type crossing there.

Step 3: Symmetry

$f(-x) = x^4 + 4x^3 \neq f(x)$ and $\neq -f(x)$. No symmetry.

Step 4: End Behavior

As $x \to \pm\infty$, $f(x) \to +\infty$ (positive leading coefficient, even degree).

Step 5: First Derivative Analysis

$$f’(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$$

Critical points: $x = 0$ and $x = 3$.

Sign analysis:

  • For $x < 0$: $4x^2 > 0$ and $(x-3) < 0$, so $f’(x) < 0$ (decreasing)
  • For $0 < x < 3$: $4x^2 > 0$ and $(x-3) < 0$, so $f’(x) < 0$ (decreasing)
  • For $x > 3$: $4x^2 > 0$ and $(x-3) > 0$, so $f’(x) > 0$ (increasing)

Step 6: Local Extrema

At $x = 0$: $f’$ does not change sign (negative on both sides). Not a local extremum.

At $x = 3$: $f’$ changes from negative to positive. Local minimum. $f(3) = 81 - 108 = -27$. Local min at $(3, -27)$.

Step 7: Second Derivative Analysis

$$f’’(x) = 12x^2 - 24x = 12x(x - 2)$$

$f’’(x) = 0$ when $x = 0$ or $x = 2$.

Sign analysis:

  • For $x < 0$: $12x < 0$ and $(x-2) < 0$, so $f’’(x) > 0$ (concave up)
  • For $0 < x < 2$: $12x > 0$ and $(x-2) < 0$, so $f’’(x) < 0$ (concave down)
  • For $x > 2$: $12x > 0$ and $(x-2) > 0$, so $f’’(x) > 0$ (concave up)

Step 8: Inflection Points

At $x = 0$: Concavity changes from up to down. Inflection point: $(0, 0)$

At $x = 2$: Concavity changes from down to up. $f(2) = 16 - 32 = -16$. Inflection point: $(2, -16)$

Step 9: Summary Table

Feature Location Value
$y$-intercept $x = 0$ $(0, 0)$
$x$-intercepts $x = 0, 4$ $(0, 0)$, $(4, 0)$
Local minimum $x = 3$ $(3, -27)$
Inflection points $x = 0, 2$ $(0, 0)$, $(2, -16)$
Increasing $x > 3$
Decreasing $x < 3$
Concave up $x < 0$ or $x > 2$
Concave down $0 < x < 2$

Step 10: Sketch Description

Starting from the upper left (as $x \to -\infty$, $y \to +\infty$), the curve descends while concave up. It passes through the origin, which is both an $x$-intercept and an inflection point. After the origin, the curve continues downward but is now concave down. At $(2, -16)$, it switches back to concave up while still decreasing. The curve reaches its lowest point at the local minimum $(3, -27)$, then increases, crossing the $x$-axis at $(4, 0)$ and continuing upward toward infinity.

Example 6: Function with Vertical Asymptote

Problem: Sketch $f(x) = \frac{x^2}{x - 1}$.

Solution:

Step 1: Domain

All real numbers except $x = 1$: $(-\infty, 1) \cup (1, \infty)$.

Step 2: Intercepts

$y$-intercept: $f(0) = \frac{0}{-1} = 0$. Point: $(0, 0)$

$x$-intercepts: $\frac{x^2}{x-1} = 0$ when $x^2 = 0$, so $x = 0$. Point: $(0, 0)$

Step 3: Symmetry

$f(-x) = \frac{x^2}{-x-1} \neq f(x)$ and $\neq -f(x)$. No symmetry.

Step 4: Asymptotes

Vertical: At $x = 1$: $$\lim_{x \to 1^+} \frac{x^2}{x-1} = \frac{1}{0^+} = +\infty$$ $$\lim_{x \to 1^-} \frac{x^2}{x-1} = \frac{1}{0^-} = -\infty$$

Slant asymptote: Since degree of numerator exceeds denominator by 1, perform long division: $$\frac{x^2}{x-1} = x + 1 + \frac{1}{x-1}$$

Slant asymptote: $y = x + 1$

Step 5: First Derivative Analysis

$$f’(x) = \frac{2x(x-1) - x^2(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}$$

Critical points: $x = 0$ and $x = 2$ (note: $x = 1$ is not in the domain).

Sign analysis:

  • For $x < 0$: $\frac{(-)(-)}{(+)} > 0$ (increasing)
  • For $0 < x < 1$: $\frac{(+)(-)}{(+)} < 0$ (decreasing)
  • For $1 < x < 2$: $\frac{(+)(-)}{(+)} < 0$ (decreasing)
  • For $x > 2$: $\frac{(+)(+)}{(+)} > 0$ (increasing)

Step 6: Local Extrema

At $x = 0$: $f’$ changes from $+$ to $-$. Local maximum at $(0, 0)$.

At $x = 2$: $f’$ changes from $-$ to $+$. Local minimum at $(2, f(2)) = (2, 4)$.

Step 7: Second Derivative (for concavity)

After calculation: $$f’’(x) = \frac{2}{(x-1)^3}$$

  • For $x < 1$: $(x-1)^3 < 0$, so $f’’(x) < 0$ (concave down)
  • For $x > 1$: $(x-1)^3 > 0$, so $f’’(x) > 0$ (concave up)

No inflection points (concavity changes at the vertical asymptote, not at a point on the curve).

Step 8: Sketch Description

Left branch ($x < 1$): Comes from above the slant asymptote, has a local max at the origin, then plunges to $-\infty$ as $x \to 1^-$. Always concave down.

Right branch ($x > 1$): Rises from $+\infty$ as $x \to 1^+$, decreases to a local min at $(2, 4)$, then increases toward the slant asymptote $y = x + 1$. Always concave up.

Key Properties and Rules

The First Derivative Test

For a critical point at $x = c$:

Sign change of $f'$ Conclusion
$+$ to $-$ Local maximum at $x = c$
$-$ to $+$ Local minimum at $x = c$
No sign change Neither max nor min

The Second Derivative Test

For a critical point at $x = c$ where $f’(c) = 0$:

Value of $f’’(c)$ Conclusion
$f’’(c) > 0$ Local minimum (concave up)
$f’’(c) < 0$ Local maximum (concave down)
$f’’(c) = 0$ Test inconclusive

Relationship Between Function, First, and Second Derivatives

$f(x)$ $f’(x)$ $f’’(x)$
Increasing Positive
Decreasing Negative
Local max/min Zero or undefined
Concave up Increasing Positive
Concave down Decreasing Negative
Inflection point Local max/min Zero (with sign change)

Quick Reference for Rational Functions

For $f(x) = \frac{p(x)}{q(x)}$ where $p$ and $q$ are polynomials:

Vertical asymptotes: Where $q(x) = 0$ and $p(x) \neq 0$

Horizontal asymptotes:

  • deg$(p)$ < deg$(q)$: $y = 0$
  • deg$(p)$ = deg$(q)$: $y = \frac{\text{leading coeff of } p}{\text{leading coeff of } q}$
  • deg$(p)$ > deg$(q)$: None (check for slant)

Holes: Where both $p(x) = 0$ and $q(x) = 0$ (common factors)

Real-World Applications

Economic Analysis: Supply and Demand Curves

Economists use curve sketching to understand market equilibrium. Supply curves typically increase (producers supply more at higher prices), while demand curves decrease (consumers buy less at higher prices). The shapes of these curves – whether they are concave up or down – reveal important information about price elasticity and market sensitivity.

When analyzing a cost function $C(x)$, the first derivative $C’(x)$ gives the marginal cost (cost of producing one more unit). Finding where the marginal cost curve has a minimum tells managers the most efficient production level.

Population Dynamics

Population growth often follows an S-shaped (logistic) curve. Initially, growth is slow (concave up), then accelerates through an inflection point, and finally slows as the population approaches its carrying capacity (concave down). Identifying the inflection point tells biologists when the population growth rate is at its maximum – crucial information for conservation efforts.

The logistic model is: $$P(t) = \frac{K}{1 + Ae^{-rt}}$$

where $K$ is the carrying capacity. The inflection point occurs at $P = K/2$, when the population is at half its maximum sustainable level.

Marginal Analysis in Business

In business, “marginal” refers to the rate of change. Curve sketching helps answer questions like:

  • At what production level does profit stop increasing? (Find where $P’(x) = 0$)
  • When does increasing production become less efficient? (Find the inflection point of $P(x)$)
  • What is the optimal price to maximize revenue? (Find the critical point of $R(p)$)

A company might find that revenue $R(x) = 100x - 0.5x^2$ has maximum at $x = 100$ units. The second derivative $R’’(x) = -1 < 0$ confirms this is a maximum, not a minimum.

Engineering: Stress-Strain Curves

When engineers test materials, they plot stress (force per area) versus strain (deformation). The shape of this curve reveals:

  • The elastic region (linear portion where the material returns to original shape)
  • The yield point (where permanent deformation begins)
  • Ultimate tensile strength (maximum stress the material can withstand)

Understanding concavity helps engineers predict material failure. A curve that becomes concave down after the yield point indicates the material is weakening and approaching failure.

Physics: Motion Analysis

In physics, position, velocity, and acceleration are related by derivatives:

  • Position: $s(t)$
  • Velocity: $v(t) = s’(t)$
  • Acceleration: $a(t) = s’’(t) = v’(t)$

Sketching a position curve lets you visualize motion. When $s’’(t) > 0$, the object is speeding up (if moving in positive direction) or slowing down (if moving in negative direction). Inflection points in the position curve correspond to maximum or minimum velocity.

Self-Test Problems

Problem 1: For $f(x) = x^3 - 6x^2 + 9x$, find all critical points and classify each as a local maximum, local minimum, or neither.

Show Answer

First, find the derivative: $$f’(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$$

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

Sign analysis:

  • For $x < 1$: $f’(x) = 3(-)(-) > 0$
  • For $1 < x < 3$: $f’(x) = 3(+)(-) < 0$
  • For $x > 3$: $f’(x) = 3(+)(+) > 0$

At $x = 1$: $f’$ changes from $+$ to $-$, so local maximum at $(1, f(1)) = (1, 4)$

At $x = 3$: $f’$ changes from $-$ to $+$, so local minimum at $(3, f(3)) = (3, 0)$

Problem 2: Find the intervals where $f(x) = x^4 - 2x^2$ is concave up and concave down.

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Find the second derivative: $$f’(x) = 4x^3 - 4x$$ $$f’’(x) = 12x^2 - 4 = 4(3x^2 - 1)$$

Set $f’’(x) = 0$: $$3x^2 - 1 = 0 \Rightarrow x^2 = \frac{1}{3} \Rightarrow x = \pm\frac{1}{\sqrt{3}}$$

Sign analysis:

  • For $x < -\frac{1}{\sqrt{3}}$: $f’’(x) > 0$ (concave up)
  • For $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$: $f’’(x) < 0$ (concave down)
  • For $x > \frac{1}{\sqrt{3}}$: $f’’(x) > 0$ (concave up)

Concave up: $\left(-\infty, -\frac{1}{\sqrt{3}}\right) \cup \left(\frac{1}{\sqrt{3}}, \infty\right)$

Concave down: $\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$

Problem 3: Find all asymptotes of $f(x) = \frac{2x^2 - 3x + 1}{x - 2}$.

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Vertical asymptote: The denominator equals zero when $x = 2$. Check that the numerator is not zero at $x = 2$: $2(4) - 3(2) + 1 = 8 - 6 + 1 = 3 \neq 0$.

Vertical asymptote: $x = 2$

Slant asymptote: Since degree of numerator (2) exceeds degree of denominator (1) by exactly 1, perform polynomial long division:

$$\frac{2x^2 - 3x + 1}{x - 2} = 2x + 1 + \frac{3}{x-2}$$

Slant asymptote: $y = 2x + 1$

No horizontal asymptote (the slant asymptote replaces it).

Problem 4: Determine whether $f(x) = \frac{x^3}{x^2 + 1}$ is even, odd, or neither.

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Calculate $f(-x)$: $$f(-x) = \frac{(-x)^3}{(-x)^2 + 1} = \frac{-x^3}{x^2 + 1}$$

Compare with $-f(x)$: $$-f(x) = -\frac{x^3}{x^2 + 1} = \frac{-x^3}{x^2 + 1}$$

Since $f(-x) = -f(x)$, the function is odd.

This means the graph has origin symmetry – if you rotate it 180 degrees about the origin, it looks the same.

Problem 5: For $f(x) = xe^{-x}$, find the inflection point(s).

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Find the derivatives using the product rule: $$f’(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$$ $$f’’(x) = -e^{-x}(1-x) + e^{-x}(-1) = e^{-x}(-1 + x - 1) = e^{-x}(x - 2)$$

Set $f’’(x) = 0$: $$e^{-x}(x - 2) = 0$$

Since $e^{-x} > 0$ always, we need $x - 2 = 0$, so $x = 2$.

Check sign change:

  • For $x < 2$: $f’’(x) = e^{-x}(\text{negative}) < 0$
  • For $x > 2$: $f’’(x) = e^{-x}(\text{positive}) > 0$

Concavity changes at $x = 2$.

Calculate $f(2) = 2e^{-2} = \frac{2}{e^2}$

Inflection point: $\left(2, \frac{2}{e^2}\right) \approx (2, 0.271)$

Problem 6: Sketch $f(x) = \frac{1}{x^2 - 4}$ by identifying all key features.

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Domain: All $x$ except where $x^2 - 4 = 0$, i.e., $x \neq \pm 2$

Intercepts:

  • $y$-intercept: $f(0) = \frac{1}{-4} = -\frac{1}{4}$
  • $x$-intercepts: None (numerator is never 0)

Symmetry: $f(-x) = \frac{1}{(-x)^2 - 4} = \frac{1}{x^2 - 4} = f(x)$. The function is even.

Asymptotes:

  • Vertical: $x = -2$ and $x = 2$
  • Horizontal: $\lim_{x \to \pm\infty} \frac{1}{x^2-4} = 0$, so $y = 0$

First derivative: $$f’(x) = \frac{-2x}{(x^2-4)^2}$$

Critical point at $x = 0$ (and undefined at $x = \pm 2$)

  • For $x < -2$: $f’(x) > 0$ (increasing)
  • For $-2 < x < 0$: $f’(x) > 0$ (increasing)
  • For $0 < x < 2$: $f’(x) < 0$ (decreasing)
  • For $x > 2$: $f’(x) < 0$ (decreasing)

Local maximum at $x = 0$: $(0, -\frac{1}{4})$

Behavior near asymptotes:

  • As $x \to 2^+$: $f(x) \to +\infty$
  • As $x \to 2^-$: $f(x) \to -\infty$
  • As $x \to -2^+$: $f(x) \to -\infty$
  • As $x \to -2^-$: $f(x) \to +\infty$

Sketch: Three branches: one for $x < -2$ (above asymptote, decreasing toward 0), one for $-2 < x < 2$ (below the $x$-axis with max at $(0, -1/4)$), and one for $x > 2$ (above asymptote, approaching 0).

Summary

  • Domain and intercepts establish where your function exists and crosses the axes, giving you anchor points for your sketch.

  • Symmetry can cut your work in half: even functions reflect across the $y$-axis, odd functions have origin symmetry.

  • Asymptotes describe behavior at extremes: vertical asymptotes where the function blows up, horizontal asymptotes where it levels off.

  • The first derivative $f’(x)$ reveals where the function increases ($f’ > 0$) and decreases ($f’ < 0$). Critical points occur where $f’ = 0$ or is undefined.

  • Local extrema occur at critical points where the derivative changes sign. Use the first or second derivative test to classify them.

  • The second derivative $f’’(x)$ determines concavity: concave up ($f’’ > 0$) looks like a cup, concave down ($f’’ < 0$) looks like a cap.

  • Inflection points occur where concavity changes, requiring $f’’ = 0$ (or undefined) with a sign change.

  • The systematic checklist (domain, intercepts, symmetry, asymptotes, first derivative, second derivative, key points, sketch) works for any function.

  • Curve sketching connects calculus to real applications: economics, biology, engineering, and physics all rely on understanding how functions behave.

  • With practice, you will develop intuition for function behavior and be able to anticipate the shape of a curve before you even start calculating.