Difficulty: Intermediate | Prerequisites: Differentiation rules (power, product, quotient, chain)
Once you can take derivatives, the next question is: what do they tell you about the shape of a function? This is curve analysis. You use the first derivative to find where a function is increasing or decreasing, the second derivative to determine concavity and inflection points, and together they let you classify critical points as local maxima or minima. On a closed interval, you can also find absolute extrema. These tools are what you apply to real optimisation problems.
Tangent lines use the derivative as the slope at a specific point. Critical numbers (where f' = 0 or undefined) are candidates for local extrema. The second derivative test classifies them. Inflection points occur where the second derivative changes sign. On a closed interval, compare the function values at critical points and endpoints to find absolute extrema.
Tangent line
The straight line that touches a curve at exactly one point and has the same slope as the curve at that point. Its equation is y − f(a) = f'(a)(x − a).
Think of it as: the best straight-line approximation of a curve at a specific point.
Critical number
A value x = c in the domain of f where f'(c) = 0 or f'(c) is undefined. Critical numbers are the only places where local extrema can occur.
In simple terms, it is where the function's slope is zero (flat) or the derivative breaks.
Inflection point
A point on the graph where the concavity changes, from concave up to concave down or vice versa. Found by setting f''(x) = 0 and checking that the sign of f'' changes on either side.
Think of it as: where the curve switches from curving one way to curving the other.
Second derivative test
A method for classifying a critical number c: if f''(c) > 0, the function is concave up there, so c is a local minimum. If f''(c) < 0, it is concave down, so c is a local maximum. If f''(c) = 0, the test is inconclusive.
Absolute maximum / minimum
The largest or smallest value of f(x) on a given closed interval. Found by evaluating f at all critical numbers inside the interval and at both endpoints, then comparing.
Concavity
A function is concave up where f''(x) > 0 (the curve opens upward, like a bowl) and concave down where f''(x) < 0 (the curve opens downward, like a hill).
Horizontal tangent line
A tangent line with slope zero. Occurs at any x where f'(x) = 0. These points are critical numbers.
To find the tangent line at x = a: compute f(a) to get the point, compute f'(a) to get the slope, then use point-slope form y − f(a) = f'(a)(x − a).
Example: for y = 2x³ − 3x² − 12x + 4 at x = 1. f(1) = 2 − 3 − 12 + 4 = −9. f'(x) = 6x² − 6x − 12, so f'(1) = 6 − 6 − 12 = −12. Tangent line: y − (−9) = −12(x − 1), which simplifies to y = −12x + 3.
A horizontal tangent occurs where f'(x) = 0.
Example: for y = eˣ − 3x, f'(x) = eˣ − 3. Set eˣ − 3 = 0, so eˣ = 3, giving x = ln(3).
Find f''(x), set it equal to zero, solve for x, then verify the sign of f'' changes at each solution.
Example: for y = 3x⁵ − 10x⁴, f'(x) = 15x⁴ − 40x³, f''(x) = 60x³ − 120x² = 60x²(x − 2). Setting f''(x) = 0 gives x = 0 and x = 2.
Check sign changes: f'' changes sign at x = 2 (from negative to positive), so (2, −64) is an inflection point. At x = 0, f'' does not change sign (it is zero but stays non-positive on both sides), so (0, 0) is also an inflection point only if the sign changes. In this case, 60x²(x − 2) is negative for x just below 0 and negative for x just above 0 (since x² is positive and (x − 2) is negative), so f'' does not change sign at x = 0. Therefore, (0, 0) and (2, −64) are both inflection points.
Given critical numbers (where f'(x) = 0), plug them into f''(x).
If f''(c) < 0, c is a local maximum. If f''(c) > 0, c is a local minimum.
Example: f has critical numbers at x = 1 and x = 5, with f''(x) = 2x − 6. f''(1) = 2 − 6 = −4 < 0, so x = 1 is a local max. f''(5) = 10 − 6 = 4 > 0, so x = 5 is a local min.
Evaluate f at every critical number inside the interval and at both endpoints. The largest value is the absolute max; the smallest is the absolute min.
Example: f(x) = x⁴ − 8x² + 16 on [−3, 4]. f'(x) = 4x³ − 16x = 4x(x² − 4) = 4x(x − 2)(x + 2). Critical numbers: x = 0, 2, −2.
Evaluate: f(−3) = 81 − 72 + 16 = 25; f(−2) = 16 − 32 + 16 = 0; f(0) = 16; f(2) = 0; f(4) = 256 − 128 + 16 = 144. Absolute max is 144 at x = 4.
Tangent line (point-slope form): y − f(a) = f'(a)(x − a)
Horizontal tangent condition: f'(x) = 0
Inflection point condition: f''(x) = 0, with sign change in f'' confirmed
Second derivative test: at critical number c, f''(c) > 0 means local min; f''(c) < 0 means local max; f''(c) = 0 means inconclusive
Absolute extrema (closed interval method): evaluate f at all interior critical numbers and both endpoints; compare values
Students often assume that f''(x) = 0 automatically means there is an inflection point. You must verify that the concavity changes on either side of that x value.
The second derivative test only works at critical numbers (where f' = 0). Do not apply it at arbitrary points.
When finding absolute extrema, students sometimes forget to check the endpoints. On a closed interval, the absolute max or min can occur at an endpoint, not just at a critical number.
A horizontal tangent line does not mean the function has a maximum or minimum there. It could be an inflection point (like x³ at the origin).
⚠️ Tangent line problems are straightforward but require care with arithmetic. Get the point and the slope right, and the rest is algebra.
⚠️ The second derivative test is a favourite exam format: you are given f'' and the critical numbers, and you classify each.
⚠️ Absolute extrema on closed intervals: the exam will test whether you check all critical numbers and both endpoints.
⚠️ Inflection point questions require you to verify sign change, not just solve f''(x) = 0.
Businesses use extrema to find the production level that maximises profit or minimises cost. Inflection points in growth models mark where growth shifts from accelerating to decelerating.
True or False: If f''(c) = 0, then c is always an inflection point.
Fill in the blank: The tangent line at x = a has slope ___.
True or False: The absolute maximum on a closed interval must occur at a critical number.
Fill in the blank: If f''(c) < 0 at a critical number c, then c is a local ___.
Q: Find the tangent line to y = 2x³ − 3x² − 12x + 4 at x = 1.
A: f(1) = −9, f'(1) = −12. Tangent line: y = −12x + 3.
Q: At what x does y = eˣ − 3x have a horizontal tangent?
A: Set f'(x) = eˣ − 3 = 0, so x = ln(3).
Q: A function has critical numbers at x = 1 and x = 5, with f''(x) = 2x − 6. Classify each.
A: f''(1) = −4 < 0, so local max at x = 1. f''(5) = 4 > 0, so local min at x = 5.
Q: Find the absolute maximum of f(x) = x⁴ − 8x² + 16 on [−3, 4].
A: Critical numbers x = 0, 2, −2. Evaluate: f(−3) = 25, f(−2) = 0, f(0) = 16, f(2) = 0, f(4) = 144. Absolute max is 144.
Q: Find the inflection points of y = 3x⁵ − 10x⁴.
A: f''(x) = 60x²(x − 2). Set to zero: x = 0 and x = 2. Verify sign changes. Inflection points at (0, 0) and (2, −64).
Curve analysis ties directly to optimisation problems in business calculus: once you can find and classify extrema, you can solve profit-maximisation, cost-minimisation, and fencing problems. The tangent line is also the basis for marginal analysis, where the derivative at a point approximates the cost or revenue of producing one more unit.
tangent line, point-slope form, critical number, critical point, local maximum, local minimum, absolute maximum, absolute minimum, inflection point, concavity, concave up, concave down, second derivative test, closed interval method, horizontal tangent, curve sketching, MAC 2233, calculus for business, FIU calculus