Difficulty: Intermediate
Prerequisites: Sections 3.1 through 3.4 (critical points, first derivative test, concavity, inflection points). You need to be comfortable finding derivatives using the power, product, quotient and chain rules before any of this will land.
This material sits near the end of the "Applications of Differentiation" chapter, which is really about turning raw derivative calculations into usable information about a function's shape. You have already learned to find critical points and to classify them with sign charts (the first derivative test). The second derivative test gives you a shortcut: if the second derivative exists and is nonzero at a critical point, one number tells you whether the point is a local max or local min, no sign chart needed. The trade-off is that the test sometimes fails (when the second derivative is zero), in which case you fall back to the first derivative test.
The second derivative test classifies a critical point as a local min or local max by checking the sign of f''(c). Positive means concave up, so a local min; negative means concave down, so a local max. If f''(c) = 0, the test is inconclusive and you revert to the first derivative test.
Critical point
A value c in the domain of f where f'(c) = 0 or f'(c) does not exist. In simple terms, these are the x-values where the slope is flat or undefined, the only places a local max or min can occur.
Second derivative test
A method for classifying a critical point c (where f'(c) = 0) by evaluating f''(c). If f''(c) > 0 the point is a local minimum; if f''(c) < 0 it is a local maximum; if f''(c) = 0 the test is inconclusive. Think of it as asking: is the curve bending upward (bowl) or downward (hill) at that critical point?
Concave up
A function is concave up on an interval when f''(x) > 0 there. The graph curves like a bowl or a smile. In simple terms, the slope is increasing as you move left to right.
Concave down
A function is concave down on an interval when f''(x) < 0 there. The graph curves like a hill or a frown. In simple terms, the slope is decreasing as you move left to right.
Inflection point
A point on the graph where the concavity changes from up to down or from down to up. This is where f''(x) = 0 or f''(x) does not exist, and the sign of f'' switches. Think of it as the moment the curve stops bending one way and starts bending the other.
Local minimum
A point where f(c) is less than or equal to f(x) for all x near c. The function dips to a low point and then rises again. At a local min with f'(c) = 0, the second derivative test says f''(c) > 0.
Local maximum
A point where f(c) is greater than or equal to f(x) for all x near c. The function rises to a high point and then falls again. At a local max with f'(c) = 0, the second derivative test says f''(c) < 0.
Increasing / decreasing
A function is increasing on an interval when f'(x) > 0 for all x in that interval, and decreasing when f'(x) < 0. These intervals are found by solving f'(x) = 0, locating where f' is undefined, and testing the sign of f' in each sub-interval.
Suppose f'' is continuous near c.
(a) If f'(c) = 0 and f''(c) > 0, then f has a local minimum at c.
(b) If f'(c) = 0 and f''(c) < 0, then f has a local maximum at c.
If f'(c) = 0 and f''(c) = 0, the test is inconclusive. In such cases, we must use the first derivative test.
Find f'(x) and set f'(x) = 0 to locate the critical points.
Find f''(x).
Plug each critical point c into f''(x).
Read the sign:
f''(c) > 0 means the graph is concave up at c, so c is a local min.
f''(c) < 0 means the graph is concave down at c, so c is a local max.
f''(c) = 0 means the test tells you nothing. Go back to a sign chart of f'.
The first derivative test never fails: it works on any critical point, including where the derivative does not exist. The second derivative test is faster when f'' is easy to compute and nonzero at the critical point. In higher dimensions (MTH 234), the second derivative test generalises and becomes the standard tool, so practising it now is worthwhile.
f'(c) = 0 \quad \text{and} \quad f''(c) > 0 \;\Rightarrow\; \text{local minimum at } cf'(c) = 0 \quad \text{and} \quad f''(c) < 0 \;\Rightarrow\; \text{local maximum at } cf'(c) = 0 \quad \text{and} \quad f''(c) = 0 \;\Rightarrow\; \text{inconclusive (use first derivative test)}Given f(x) = x^2 / (x - 1), with f'(x) = (x^2 - 2x) / (x - 1)^2 and f''(x) = 2 / (x - 1)^3.
Step 1. Find critical points: set f'(x) = 0.
x^2 - 2x = 0 gives x(x - 2) = 0, so x = 0 and x = 2. Note x = 1 is not in the domain.
Step 2. Evaluate f'' at each critical point.
f''(0) = 2 / (0 - 1)^3 = 2 / (-1) = -2. Since f''(0) < 0, there is a local maximum at x = 0.
f''(2) = 2 / (2 - 1)^3 = 2 / 1 = 2. Since f''(2) > 0, there is a local minimum at x = 2.
Find where f is increasing/decreasing, concave up/down, and identify local extrema and inflection points.
Increasing/decreasing.
f'(x) = -1/x^2 + 1/16. Set equal to zero: 1/16 = 1/x^2, so x^2 = 16, giving x = 4 and x = -4. Also note f is undefined at x = 0.
f'(x) > 0 on (-inf, -4) and (4, inf), so f is increasing there.
f'(x) < 0 on (-4, 0) and (0, 4), so f is decreasing there.
Concavity.
f''(x) = 2/x^3.
f''(x) > 0 when x > 0, so f is concave up on (0, inf).
f''(x) < 0 when x < 0, so f is concave down on (-inf, 0).
There is no inflection point because f is not defined at x = 0, so the concavity cannot switch through a point on the graph.
Local extrema.
At x = -4: f'(-4) = 0 and f''(-4) = 2/(-64) < 0, so local maximum at x = -4.
At x = 4: f'(4) = 0 and f''(4) = 2/64 > 0, so local minimum at x = 4.
Derivative. f'(x) = cos x - sin x. Set to zero: cos x = sin x, so tan x = 1, giving x = pi/4 and x = 5pi/4 on [0, 2pi].
Classify using the second derivative.
f''(x) = -sin x - cos x.
f''(pi/4) = -sin(pi/4) - cos(pi/4) = -sqrt(2)/2 - sqrt(2)/2 = -sqrt(2). Since f''(pi/4) < 0, there is a local maximum at x = pi/4, with f(pi/4) = sqrt(2).
f''(5pi/4) = -sin(5pi/4) - cos(5pi/4) = sqrt(2)/2 + sqrt(2)/2 = sqrt(2). Since f''(5pi/4) > 0, there is a local minimum at x = 5pi/4, with f(5pi/4) = -sqrt(2).
Intervals.
Increasing on (0, pi/4) and (5pi/4, 2pi).
Decreasing on (pi/4, 5pi/4).
Find intervals of increase/decrease and local extrema.
Derivative (using the product rule with fractional exponents):
f'(x) = (4 - x) / [x^(1/3) (6 - x)^(2/3)].
Critical points. f'(x) = 0 when x = 4. f'(x) is undefined at x = 0 and x = 6 (both in the domain of f).
Sign analysis of f':
f'(x) > 0 on (0, 4), so f is increasing.
f'(x) < 0 on (-inf, 0) and (4, 6) and (6, inf), so f is decreasing.
Local extrema:
At x = 0, f' changes from negative to positive, so there is a local minimum at x = 0.
At x = 4, f' changes from positive to negative, so there is a local maximum at x = 4.
Note: the second derivative test cannot be applied at x = 0 or x = 6 because f' does not exist there. At x = 4 you could use either test; the first derivative test (sign chart) works for all three critical points.
(a) Critical numbers. f'(x) = 12x^2. Setting 12x^2 = 0 gives x = 0 as the only critical number.
(b) Increasing/decreasing. f'(x) = 12x^2 >= 0 for all x, and equals zero only at x = 0.
f is increasing on (-inf, 0) and (0, inf) (or equivalently, on all of R).
f is never decreasing.
(c) Classification. f''(x) = 24x, so f''(0) = 0. The second derivative test is inconclusive. Using the first derivative test: f' does not change sign at x = 0 (it is positive on both sides), so x = 0 is neither a local max nor a local min.
This is the classic example of the second derivative test failing. The function 4x^3 - 6 has a horizontal tangent at x = 0 but no extremum there, just a point of inflection.
Determine A and B so that y = Ax^(1/9) + Bx^(-1/9) has an inflection point at (1, 2).
Condition 1: the point (1, 2) is on the curve: A(1) + B(1) = 2, so A + B = 2.
Condition 2: an inflection point at x = 1 means y''(1) = 0.
y' = (A/9) x^(-8/9) - (B/9) x^(-10/9).
y'' = -(8A/81) x^(-17/9) + (10B/81) x^(-19/9).
At x = 1: y''(1) = (-8A + 10B) / 81 = 0, so -8A + 10B = 0, giving 4A = 5B.
Solving the system A + B = 2 and 4A = 5B yields B = 8/9 and A = 10/9.
Students often think f''(c) = 0 means "no local extremum." It does not. It means the second derivative test cannot decide. The point might still be a local max, local min, or neither. You need the first derivative test to find out.
Students sometimes apply the second derivative test at points where f' does not exist (cusps, corners, vertical tangents). The test requires f'(c) = 0 as a precondition, not merely that c is a critical point.
Confusing the sign convention: f''(c) > 0 means concave up, which is a local min (a bowl), not a max. A common memory trick is "positive second derivative = smile = minimum."
Forgetting to check the domain. In Example 3.17, x = 1 is not a critical point because it is not in the domain of f. Students sometimes include it in their list of critical numbers.
The second derivative test is a staple of Calc 1 exams. Expect at least one problem asking you to classify critical points using it.
Problems where f''(c) = 0 (the inconclusive case) appear frequently, precisely because they test whether you know to fall back to the first derivative test.
Quotient-rule functions like x^2 / (x - 1) are common exam choices because they produce a clean f'' and have domain restrictions that trip students up.
Trigonometric functions on a closed interval (like sin x + cos x on [0, 2pi]) combine derivative skills with unit-circle knowledge. These show up regularly.
Fractional-exponent functions like x^(2/3)(6 - x)^(1/3) test your ability to handle critical points where f' does not exist, where the second derivative test cannot be applied.
The "find A and B" style (Example 3.22) tests the ability to set up a system of equations from calculus conditions. This type of problem appears in WeBWorK and on exams.
True or false: If f'(c) = 0 and f''(c) = 0, then c is neither a local max nor a local min. False. The test is inconclusive; c could still be a local extremum.
Fill in the blank: If f'(c) = 0 and f''(c) > 0, then f has a local ______ at c. Minimum.
True or false: The second derivative test can be used at a point where f'(c) does not exist. False. It requires f'(c) = 0.
Fill in the blank: f''(x) > 0 on an interval means the function is concave ______ there. Up.
True or false: For f(x) = 4x^3 - 6, the point x = 0 is a local minimum. False. It is a point of inflection (f' does not change sign).
Q: Let f(x) = x^3 - 3x + 1. Find the critical points and classify each using the second derivative test.
A: f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1). Critical points at x = 1 and x = -1. f''(x) = 6x. f''(1) = 6 > 0, so local min at x = 1. f''(-1) = -6 < 0, so local max at x = -1.
Q: For f(x) = x^4, apply the second derivative test at x = 0. What happens?
A: f'(x) = 4x^3, so f'(0) = 0. f''(x) = 12x^2, so f''(0) = 0. The test is inconclusive. Using the first derivative test, f' changes from negative to positive at x = 0, so it is a local minimum.
Q: Suppose f'(3) = 0 and f''(3) = -7. What can you conclude?
A: Since f'(3) = 0 and f''(3) < 0, the second derivative test tells us f has a local maximum at x = 3.
Q: Find the intervals of increase and decrease for f(x) = 1/x + x/16 and identify any local extrema.
A: f'(x) = -1/x^2 + 1/16. Setting to zero gives x = 4 and x = -4. f is increasing on (-inf, -4) and (4, inf), decreasing on (-4, 0) and (0, 4). Local max at x = -4, local min at x = 4.
Q: Determine constants A and B so that y = Ax^(1/9) + Bx^(-1/9) has an inflection point at (1, 2).
A: Two conditions: A + B = 2 (point on the curve) and -8A + 10B = 0 (y''(1) = 0). Solving gives A = 10/9 and B = 8/9.
This material connects directly to optimisation problems (Section 3.7 and beyond), where you find critical points of a function that models a real-world quantity and use the second derivative test to confirm whether you have found a maximum or minimum. The second derivative test is also the foundation for the multivariable second derivative test in MTH 234 (Calculus III), where the test uses a determinant of a matrix of partial derivatives (the Hessian) rather than a single number.
Concavity and inflection points connect to curve sketching, where you combine information from f, f' and f'' to draw a complete picture of the graph. Understanding concavity also matters for approximation: a function that is concave up on an interval means the tangent-line approximation underestimates the true value there.
Second derivative test, Calculus 1, MTH 132, MSU, applications of differentiation, critical points, local maximum, local minimum, concavity, concave up, concave down, inflection point, first derivative test vs second derivative test, sign chart, quotient rule, product rule, fractional exponents, curve sketching, increasing and decreasing functions, Calc 1 exam review, Michigan State University