Difficulty: Intermediate | Prerequisites: Sections 1 and 2 of this chapter (critical numbers, the MVT), plus comfort with differentiation rules including the quotient rule and chain rule.
This chunk of Chapter 3 is about reading a function's behaviour from its derivatives and then putting that information together into a complete graph. The first derivative tells you where a function rises and falls; the second derivative tells you how the curve bends. Limits at infinity let you describe end behaviour, including horizontal and slant asymptotes. Curve sketching ties all of this, plus your algebra skills with intercepts, symmetry and domain, into a systematic procedure that produces an accurate graph without a calculator. This is one of the most exam-heavy parts of Calculus I.
f'(x) > 0 means increasing, f'(x) < 0 means decreasing. f''(x) > 0 means concave up, f''(x) < 0 means concave down. The first derivative test classifies critical numbers by sign changes in f'; the second derivative test does it by checking the sign of f'' at the critical number. For end behaviour, compare the degrees of numerator and denominator to find horizontal or slant asymptotes, then combine everything into a curve sketch.
Increasing function
f is increasing on (a, b) if f'(x) > 0 for all x in (a, b). The graph climbs from left to right.
Decreasing function
f is decreasing on (a, b) if f'(x) < 0 for all x in (a, b). The graph falls from left to right.
Concave up
The graph of f lies above all of its tangent lines on an interval. Equivalently, f''(x) > 0 on that interval. Think of it as a cup that holds water.
Concave down
The graph of f lies below all of its tangent lines on an interval. Equivalently, f''(x) < 0 on that interval. Think of it as a cap or a frown.
Inflection point
A point where the curve changes concavity (from concave up to concave down, or vice versa). f must be continuous there, and f'' = 0 or f'' undefined is a necessary condition, though not sufficient.
First Derivative Test
A method for classifying critical numbers by examining how f' changes sign around them. Positive-to-negative means a local max; negative-to-positive means a local min; no sign change means neither.
Second Derivative Test
A quicker classification: at a critical number c where f'(c) = 0, if f''(c) > 0 then c is a local min; if f''(c) < 0 then c is a local max; if f''(c) = 0 the test is inconclusive.
Horizontal asymptote
The line y = L is a horizontal asymptote of f if the limit of f(x) as x approaches infinity or negative infinity equals L. In simple terms, it is the height the curve levels off to far to the right or far to the left.
Slant (oblique) asymptote
A non-horizontal line y = mx + b that the graph approaches as x goes to plus or minus infinity. It occurs for a rational function when the degree of the numerator is exactly one more than the degree of the denominator. Found by polynomial long division.
Limits at infinity
The value a function approaches as x grows without bound (to infinity) or decreases without bound (to negative infinity). The key technique: divide numerator and denominator by the highest power of x in the denominator.
Even function
f(-x) = f(x) for all x. The graph is symmetric about the y-axis.
Odd function
f(-x) = -f(x) for all x. The graph is symmetric about the origin.
Find f'(x). Set f'(x) = 0 and find where f'(x) is undefined to get partition points.
Build a sign chart for f'(x) over the resulting intervals.
f'(x) > 0 on an interval means f is increasing there; f'(x) < 0 means decreasing.
Example: f(x) = x^2 + 4x + 5. f'(x) = 2x + 4 = 0 at x = -2. For x < -2, f'(x) < 0 (decreasing). For x > -2, f'(x) > 0 (increasing).
Find f''(x). Set f''(x) = 0 and find where f''(x) is undefined to get partition points.
Build a sign chart for f''(x).
f''(x) > 0 means concave up; f''(x) < 0 means concave down.
An inflection point occurs where the concavity changes (f'' switches sign). The function must be continuous there.
f''(x) = 0 does not guarantee an inflection point. Example: f(x) with f''(x) = (x + 1)^2(x + 5). f'' = 0 at x = -1 and x = -5. At x = -1, (x + 1)^2 is always non-negative so f'' does not change sign. No inflection point at x = -1. At x = -5, f'' changes sign, so x = -5 is an inflection point.
At a critical number c of f:
f' changes from + to - at c: local maximum at c.
f' changes from - to + at c: local minimum at c.
f' does not change sign at c: neither a max nor a min.
Example: f(x) = 2x^3 + 3x^2 - 36x. f'(x) = 6x^2 + 6x - 36 = 6(x + 3)(x - 2). Critical numbers: x = -3 and x = 2. Sign chart: f' > 0 on (-inf, -3), f' < 0 on (-3, 2), f' > 0 on (2, inf). So x = -3 is a local max and x = 2 is a local min.
At a critical number c where f'(c) = 0:
f''(c) > 0: local minimum (curve is concave up, so the flat point is a valley).
f''(c) < 0: local maximum (curve is concave down, so the flat point is a hilltop).
f''(c) = 0: the test fails; revert to the first derivative test.
Example: f(x) = x^2 / (x - 1). f'(x) = (x^2 - 2x) / (x - 1)^2 = x(x - 2) / (x - 1)^2. Critical numbers: x = 0 and x = 2. f''(x) = 2 / (x - 1)^3. f''(0) = 2 / (-1) = -2 < 0, so x = 0 is a local max. f''(2) = 2 / 1 = 2 > 0, so x = 2 is a local min.
For f(x) = p(x)/q(x) where p and q are polynomials with leading coefficients a and b:
deg(p) > deg(q): no horizontal asymptote.
deg(p) < deg(q): horizontal asymptote y = 0.
deg(p) = deg(q): horizontal asymptote y = a/b.
Divide every term in the numerator and denominator by the highest power of x in the denominator. As x approaches infinity, terms like 1/x, 1/x^2, etc. approach 0.
Example: lim (x -> inf) (2x - 1) / (5x + 3). Divide by x: (2 - 1/x) / (5 + 3/x) -> 2/5.
When taking the limit as x -> -inf of a function with sqrt(x^2), remember that sqrt(x^2) = |x| = -x when x is negative. This can produce different horizontal asymptotes in each direction.
Occur when deg(numerator) = deg(denominator) + 1. Found by polynomial long division: the quotient is the slant asymptote, and the remainder over the divisor goes to 0 as x -> inf.
Example: y = (2x^2 - x) / (x + 1). Long division gives 2x - 3 + 3/(x + 1). The slant asymptote is y = 2x - 3.
Example: y = (x^2 + 3x + 2) / (x - 2). Long division gives x + 5 + 12/(x - 2). The slant asymptote is y = x + 5.
If r > 0, then lim (x -> inf) 1/x^r = 0. If r > 0 and x^r is defined for all x, then lim (x -> -inf) 1/x^r = 0.
For a full curve sketch, work through these steps systematically:
Domain. What x-values are allowed? Watch for division by zero and even roots of negatives.
Intercepts. y-intercept: evaluate f(0). x-intercepts: solve f(x) = 0.
Symmetry. Is f even (symmetric about y-axis), odd (symmetric about origin), or neither? Check f(-x).
Asymptotes. Vertical: where the denominator is zero (and the numerator is not). Horizontal: use the degree rules or limit calculations. Slant: polynomial long division when deg(num) = deg(den) + 1.
Increasing/decreasing. Find f'(x), its zeros, and where it is undefined. Build a sign chart.
Local extrema. Apply the first derivative test (or second derivative test) at each critical number.
Concavity. Find f''(x), its zeros, and where it is undefined. Build a sign chart.
Inflection points. Where f'' changes sign.
Sketch. Plot the key points (intercepts, extrema, inflection points), draw the asymptotes, and connect the dots respecting the increasing/decreasing and concavity information.
Domain: all real numbers except x = 1 and x = -1.
y-intercept: f(0) = 0. x-intercept: x = 0.
Symmetry: f(-x) = 2(-x)^2 / ((-x)^2 - 1) = 2x^2 / (x^2 - 1) = f(x). So f is even.
Vertical asymptotes: x = 1 and x = -1. Horizontal asymptote: deg equal, leading coefficients 2/1, so y = 2.
f'(x) = -4x / (x^2 - 1)^2. Critical number: x = 0. f'(x) > 0 for x < 0 (on the intervals where f is defined), f'(x) < 0 for x > 0. So x = 0 is a local max with f(0) = 0.
Concavity analysis via f''(x) determines the bend of the curve on each interval between asymptotes.
f'(x) > 0 \Rightarrow f \text{ increasing}; \quad f'(x) < 0 \Rightarrow f \text{ decreasing}f''(x) > 0 \Rightarrow \text{concave up}; \quad f''(x) < 0 \Rightarrow \text{concave down}\text{Horizontal asymptote for } \frac{p(x)}{q(x)}: \quad \deg p = \deg q \Rightarrow y = \frac{a}{b}\text{Slant asymptote: perform } \frac{p(x)}{q(x)} = (mx+b) + \frac{r(x)}{q(x)}, \quad \text{asymptote is } y = mx+bStudents often think f''(c) = 0 automatically means c is an inflection point. It does not. The second derivative must actually change sign at c.
Students confuse "the second derivative test fails" with "there is no extremum." When f''(c) = 0, the test is simply inconclusive; you must use the first derivative test instead.
Students assume a function cannot cross a horizontal asymptote. It can. A horizontal asymptote describes end behaviour; the function may cross y = L at finite x-values.
When evaluating limits at negative infinity involving square roots, students forget that sqrt(x^2) = -x when x < 0. This causes sign errors.
Full curve sketching problems (domain through sketch) are standard exam questions worth many points.
Expect a problem where you must find intervals of increase/decrease and classify critical numbers.
Limits at infinity with radicals (sqrt in numerator or denominator) are common tricky problems.
Slant asymptotes via long division appear regularly. Make sure you can do polynomial long division by hand.
True or false: If f'(x) > 0 on an interval, then f is concave up on that interval.
Answer: False. f'(x) > 0 means increasing. Concave up requires f''(x) > 0.
Fill in the blank: An inflection point is where the function changes ______.
Answer: concavity.
True or false: A rational function where deg(numerator) = deg(denominator) + 2 has a slant asymptote.
Answer: False. The degree difference must be exactly 1 for a slant asymptote.
Fill in the blank: If f'(c) = 0 and f''(c) > 0, then c is a local ______.
Answer: minimum.
Q: Find where f(x) = sin(x) + cos(x) on [0, 2pi] is increasing and decreasing. Identify local extrema.
A: f'(x) = cos(x) - sin(x) = 0 when tan(x) = 1, so x = pi/4 and x = 5pi/4. f' > 0 on (0, pi/4) and (5pi/4, 2pi); f' < 0 on (pi/4, 5pi/4). Local max at x = pi/4 with f(pi/4) = sqrt(2). Local min at x = 5pi/4 with f(5pi/4) = -sqrt(2).
Q: Evaluate lim (x -> inf) (3 + 2x) / (5 - x).
A: Divide numerator and denominator by x: (3/x + 2) / (5/x - 1) -> 2 / (-1) = -2. The horizontal asymptote is y = -2.
Q: Find the slant asymptote of y = (x^2 + 3x + 2) / (x - 2).
A: Polynomial long division: x^2 + 3x + 2 divided by (x - 2) gives x + 5 with remainder 12. So y = x + 5 is the slant asymptote.
Q: Determine whether g(x) = (4x^3 - x) / (x^2 + 1) is even, odd, or neither.
A: g(-x) = (4(-x)^3 - (-x)) / ((-x)^2 + 1) = (-4x^3 + x) / (x^2 + 1) = -(4x^3 - x) / (x^2 + 1) = -g(x). So g is odd.
The first derivative test relies on critical numbers from Section 1. Concavity and inflection points set the groundwork for understanding the shape of antiderivative graphs in Section 9. Curve sketching is a prerequisite skill for optimisation problems in Section 7, where you often need to confirm a critical number is a maximum or minimum. Limits at infinity connect back to the limit laws from Chapter 1.
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