MTH 132, Calculus I (Michigan State University) | Chapter 3, Sections 3–4 Source: Chapter 3 lecture notes, Applications of Differentiation
Tags: increasing, decreasing, concave up, concave down, first derivative test, second derivative test, inflection point, horizontal asymptote, limits at infinity, rational functions, Calculus I, MTH 132, MSU
Difficulty: Intermediate Prerequisites: Sections 1–2 of Chapter 3 (critical numbers, EVT, MVT), derivative rules from Chapter 2, basic inequality solving from algebra.
Sections 3 and 4 teach you to read a function's shape from its derivatives. The first derivative tells you where the function climbs and where it falls; the second derivative tells you whether the curve bends upward or downward. Together they let you classify critical points, locate inflection points, and sketch accurate graphs without a calculator. Section 4 extends the picture to the function's long-run behaviour by introducing limits at infinity and horizontal asymptotes. This is the toolkit you need for the full curve-sketching problems in Section 5.
A positive first derivative means the function is increasing; a negative one means it is decreasing. Sign changes in f' at a critical number tell you whether that point is a local max, local min, or neither. The second derivative controls concavity: positive means concave up (bowl shape), negative means concave down (hill shape). Limits at infinity describe what happens to a function's output as x grows without bound, and they determine horizontal asymptotes.
Increasing function
f is increasing on (a, b) if f'(x) > 0 for all x in (a, b). In simple terms, the curve goes uphill as you move to the right.
Decreasing function
f is decreasing on (a, b) if f'(x) < 0 for all x in (a, b). The curve goes downhill as you move to the right.
Concave up
The graph of f lies above all of its tangent lines on an interval I. Equivalently, f''(x) > 0 on I. Think of it as a bowl or a smile shape.
Concave down
The graph of f lies below all of its tangent lines on I. Equivalently, f''(x) < 0 on I. Think of it as a hill or a frown shape.
Inflection point
A point P on the curve y = f(x) where f is continuous and the concavity changes (from up to down, or down to up). Candidates occur where f''(x) = 0 or f''(x) does not exist, but a sign change in f'' must be verified.
First Derivative Test
A method for classifying a critical number c by checking whether f' changes sign at c. Positive-to-negative means local max; negative-to-positive means local min; no sign change means neither.
Second Derivative Test
If f'(c) = 0 and f''(c) > 0, then c is a local min. If f'(c) = 0 and f''(c) < 0, then c is a local max. If f''(c) = 0, the test is inconclusive and you must fall back to the First Derivative Test.
Horizontal asymptote
The line y = L is a horizontal asymptote of f if the limit of f(x) as x approaches positive or negative infinity equals L. In simple terms, it is the value the function settles toward for very large or very negative inputs.
Limit at infinity
The value that f(x) approaches as x grows without bound (x -> infinity) or decreases without bound (x -> negative infinity).
Slant (oblique) asymptote
A non-horizontal line y = mx + b that the graph approaches as x approaches infinity or negative infinity. Occurs when the degree of the numerator is exactly one more than the degree of the denominator. Found by polynomial long division.
To determine where f is increasing or decreasing, find f'(x), then solve f'(x) > 0 (increasing) and f'(x) < 0 (decreasing). Build a sign chart: identify where f'(x) = 0 or is undefined, then test one point in each sub-interval.
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).
The second derivative controls concavity. f''(x) > 0 means the curve is concave up (opening upward); f''(x) < 0 means concave down (opening downward).
Example: f(x) = x^2 + 4x + 5. f''(x) = 2 > 0 for all x, so the function is concave up everywhere. This makes sense because it is an upward-opening parabola.
Inflection points occur where f'' changes sign. Candidates: points where f''(x) = 0 or f''(x) does not exist (and f is continuous there). You must verify a sign change; f''(c) = 0 alone is not enough.
Example: suppose f''(x) = (x + 1)^2(x + 5). Setting f''(x) = 0 gives x = -1 and x = -5. Sign analysis: f'' < 0 on (-infinity, -5), f'' > 0 on (-5, -1), f'' > 0 on (-1, infinity). The sign changes only at x = -5, so the only inflection point is at x = -5. At x = -1, f'' touches zero but does not change sign, so no inflection point there.
Example: f(x) = x^3 - x^2 - 5x + 3. f''(x) = 6x - 2. Setting f''(x) = 0 gives x = 1/3. For x < 1/3, f'' < 0 (concave down). For x > 1/3, f'' > 0 (concave up). Inflection point at x = 1/3.
Suppose c is a critical number of f.
If f' changes from positive to negative at c, then f has a local maximum at c.
If f' changes from negative to positive at c, then f has a local minimum at c.
If f' does not change sign, then c is neither a local max nor a local min.
Example: f(x) = 2x^3 + 3x^2 - 36x. f'(x) = 6x^2 + 6x - 36 = 6(x^2 + x - 6) = 6(x + 3)(x - 2). Critical numbers: x = -3, x = 2. Sign chart: f' > 0 on (-infinity, -3), f' < 0 on (-3, 2), f' > 0 on (2, infinity). So x = -3 is a local max and x = 2 is a local min.
If f'(c) = 0 and f''(c) > 0, then c is a local minimum (the curve is concave up at c, so the flat tangent sits at the bottom of a bowl). If f'(c) = 0 and f''(c) < 0, then c is a local maximum. If f''(c) = 0, the test fails and you must use the First Derivative Test instead.
Example: f(x) = x^2/(x - 1). Given f'(x) = (x^2 - 2x)/(x - 1)^2 and f''(x) = 2/(x - 1)^3. Critical numbers from f'(x) = 0: x(x - 2) = 0, so x = 0 and x = 2. At x = 0: f''(0) = 2/(-1)^3 = -2 < 0, so local max. At x = 2: f''(2) = 2/(1)^3 = 2 > 0, so local min.
For f(x) = p(x)/q(x) where p and q are polynomials with leading coefficients a and b:
If deg(p) > deg(q): no horizontal asymptote (the function grows without bound).
If deg(p) < deg(q): horizontal asymptote y = 0.
If deg(p) = deg(q): horizontal asymptote y = a/b.
This works for a quick answer, but the calculus approach (dividing by the highest power of x in the denominator) handles non-polynomial cases too.
To evaluate a limit at infinity of a fraction, divide every term in both numerator and denominator by the highest power of x in the denominator. Terms like 1/x, 1/x^2, etc. all tend to zero.
Example: lim (x -> infinity) of (2x - 1)/(5x + 3). Divide top and bottom by x: (2 - 1/x)/(5 + 3/x). As x -> infinity, 1/x -> 0 and 3/x -> 0, so the limit is 2/5.
For r > 0:
lim (x -> infinity) of 1/x^r = 0.
lim (x -> negative infinity) of 1/x^r = 0 (provided x^r is defined for negative x).
When the numerator or denominator contains sqrt(x), rewrite using fractional exponents and divide by the appropriate power of x. Be careful with the sign when x -> negative infinity: sqrt(x^2) = |x| = -x when x < 0.
Example: lim (x -> infinity) of (sqrt(x) - x^3)/(2x + x^2). The dominant term in the denominator is x^2. Divide by x^2: (x^(-3/2) - x)/(2/x + 1). As x -> infinity, the numerator behaves like -x and the denominator like 1, so the limit is negative infinity (no horizontal asymptote from the right).
y = L is a horizontal asymptote of f if lim (x -> infinity) f(x) = L or lim (x -> negative infinity) f(x) = L. A function can have at most two horizontal asymptotes, one from each direction. It is a myth that a function can never cross its horizontal asymptote; many functions do (e.g. oscillating functions that settle toward a limit).
When deg(p) = deg(q) + 1 in a rational function p(x)/q(x), the function has a slant (oblique) asymptote. Find it by polynomial long division: write p(x)/q(x) = (mx + b) + r(x)/q(x). As x -> infinity, the remainder r(x)/q(x) -> 0, so y = mx + b is the slant asymptote.
Example: y = (2x^2 - x)/(x + 1). Long division gives 2x - 3 + 3/(x + 1). As x -> infinity, 3/(x + 1) -> 0, so y = 2x - 3 is the slant asymptote.
Increasing/Decreasing Test: f'(x) > 0 implies f increasing; f'(x) < 0 implies f decreasing.
Concavity Test: f''(x) > 0 implies concave up; f''(x) < 0 implies concave down.
First Derivative Test: sign change in f' at critical number c determines local max (+ to -), local min (- to +), or neither (no change).
Second Derivative Test: at a critical number where f'(c) = 0, if f''(c) > 0 then local min, if f''(c) < 0 then local max, if f''(c) = 0 then inconclusive.
Horizontal asymptote rules for p(x)/q(x):
deg(p) > deg(q): no HA.
deg(p) < deg(q): HA at y = 0.
deg(p) = deg(q): HA at y = (leading coeff of p)/(leading coeff of q).
Slant asymptote: exists when deg(p) = deg(q) + 1. Perform polynomial long division; the quotient (ignoring remainder) is the slant asymptote.
Key limit: lim (x -> infinity) 1/x^r = 0 for r > 0.
The sign of a derivative shows up everywhere a rate of change matters. In economics, a firm's revenue function is increasing where marginal revenue is positive. In biology, the concavity of a population growth curve tells you whether growth is accelerating or decelerating, and the inflection point marks the moment the growth rate peaks. Horizontal asymptotes model carrying capacity: a population that levels off at 60,000 has a horizontal asymptote at P = 60,000.
Students often assume that f''(c) = 0 automatically means c is an inflection point. It does not. The second derivative must change sign at c. For example, f(x) = x^4 has f''(0) = 0 but no inflection point at x = 0 (concave up on both sides).
Students sometimes forget that f' undefined at a point also creates a critical number. Functions with fractional exponents (like x^(2/3)) are common exam traps.
A frequent error with limits at infinity is dividing by the highest power of x in the numerator rather than the denominator. Always divide by the highest power in the denominator.
Students forget that sqrt(x^2) = |x|, not x. When x is negative, sqrt(x^2) = -x. This matters when computing limits as x -> negative infinity.
⚠️ Sign charts for f' and f'' are the backbone of almost every graph-analysis problem. Practice building them quickly.
⚠️ The Second Derivative Test is convenient but can fail (when f''(c) = 0). Always have the First Derivative Test as a backup.
⚠️ Limits at infinity with radicals are a favourite exam topic. Remember to handle the sign of x carefully when x -> negative infinity.
⚠️ Know how to do polynomial long division. Slant asymptotes show up on exams and require it.
True or False: If f'(x) > 0 on an interval, the function is concave up on that interval. Answer: False. f' > 0 means increasing. Concave up requires f'' > 0.
Fill in the blank: An inflection point is where the ____ changes sign. Answer: second derivative.
True or False: If deg(numerator) = deg(denominator), there is no horizontal asymptote. Answer: False. The horizontal asymptote is y = (leading coefficient of numerator)/(leading coefficient of denominator).
True or False: If f'(c) = 0 and f''(c) = 0, then c is an inflection point. Answer: False. The Second Derivative Test is inconclusive; c might be a local max, local min, or inflection point.
Q: Find the intervals of increase and decrease for f(x) = x/(x^2 + 1).
A: f'(x) = (1 - x^2)/(x^2 + 1)^2 using the quotient rule. f'(x) = 0 when x^2 = 1, so x = -1 and x = 1. Sign chart: f' > 0 on (-1, 1), f' < 0 on (-infinity, -1) and (1, infinity). So f is increasing on (-1, 1) and decreasing on (-infinity, -1) and (1, infinity). Local min at x = -1, local max at x = 1.
Q: Find the horizontal asymptote of y = (3x + 1)/(1 - x).
A: deg(numerator) = deg(denominator) = 1. Leading coefficients: 3 and -1. Horizontal asymptote: y = 3/(-1) = -3.
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 quotient x + 5 with remainder 12. So y = (x + 5) + 12/(x - 2). As x -> infinity, 12/(x - 2) -> 0. Slant asymptote: y = x + 5.
Q: Determine where f(x) = x^3 - x^2 - 5x + 3 has inflection points.
A: f''(x) = 6x - 2. Setting f''(x) = 0 gives x = 1/3. For x < 1/3, f'' < 0 (concave down). For x > 1/3, f'' > 0 (concave up). Sign changes at x = 1/3, so there is an inflection point at x = 1/3. f(1/3) = 1/27 - 1/9 - 5/3 + 3 = 38/27.
Sections 3 and 4 feed directly into Section 5 (Curve Sketching), where you combine domain, intercepts, symmetry, asymptotes, intervals of increase/decrease, concavity, and inflection points into a complete graph. The First and Second Derivative Tests are also essential for Section 7 (Optimisation), where you must confirm that a critical number is a max or min. Limits at infinity reappear in improper integrals in Calculus II, and the concept of horizontal asymptotes is central to the study of convergence in series.
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