MTH 132, Calculus I (Michigan State University) | Chapter 3, Sections 1–2 Source: Chapter 3 lecture notes, Applications of Differentiation
Tags: absolute maximum, absolute minimum, local max, local min, extreme values, critical number, critical point, Extreme Value Theorem, EVT, closed interval method, Rolle's Theorem, Mean Value Theorem, MVT, Calculus I, MTH 132, MSU
Difficulty: Intermediate Prerequisites: Limits and continuity (Ch. 1), derivative rules and techniques (Ch. 2). You should be comfortable differentiating polynomials, rational functions, and compositions before starting here.
Chapter 3 is where differentiation stops being an abstract skill and starts doing real work. Sections 1 and 2 introduce the idea of using derivatives to locate a function's highest and lowest values, then prove that, under the right conditions, the derivative must hit certain target slopes. These ideas underpin optimisation problems later in the chapter and recur throughout Calculus II and multivariable calculus. If you can find critical numbers, apply the Closed Interval Method, and state the Mean Value Theorem with its hypotheses, you have the foundation for nearly every application problem in this course.
Absolute and local extrema are the largest and smallest values a function takes, either globally or near a point. To find them on a closed interval, evaluate the function at its critical numbers and endpoints, then compare. The Mean Value Theorem guarantees that a smooth, continuous function must, somewhere between two endpoints, have an instantaneous rate of change equal to its average rate of change over that interval.
Absolute (global) maximum
The value f(c) is the absolute maximum of f on a domain D if f(c) >= f(x) for every x in D. In simple terms, it is the single highest output the function reaches on the entire domain.
Absolute (global) minimum
The value f(c) is the absolute minimum of f on D if f(c) <= f(x) for every x in D. Think of it as the lowest point on the entire curve over the domain.
Local (relative) maximum
f(c) is a local maximum if f(c) >= f(x) for all x near c (in some open interval around c). In simple terms, it is a peak in the function's neighbourhood, even if higher values exist elsewhere.
Local (relative) minimum
f(c) is a local minimum if f(c) <= f(x) for all x near c. Think of it as a valley in the function's immediate vicinity.
Extreme values (extrema)
A collective term for maximums and minimums, whether local or absolute.
Critical number (critical point)
A number c in the domain of f such that either f'(c) = 0 or f'(c) does not exist. These are the only interior points where extrema can occur. In simple terms, critical numbers are where the tangent line is flat or where the derivative breaks down.
Extreme Value Theorem (EVT)
If f is continuous on a closed interval [a, b], then f attains both an absolute maximum and an absolute minimum somewhere in [a, b]. Think of it as the guarantee that a continuous function on a closed interval cannot escape to infinity or leave gaps where a max or min should be.
Rolle's Theorem
If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists at least one c in (a, b) where f'(c) = 0. In simple terms, if the function starts and ends at the same height, it must have a horizontal tangent somewhere in between.
Mean Value Theorem (MVT)
If f is continuous on [a, b] and differentiable on (a, b), then there exists a c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a). Think of it as saying the instantaneous slope must match the average slope at least once.
Secant line
The straight line connecting the points (a, f(a)) and (b, f(b)). The MVT says the tangent line is parallel to the secant line at some interior point.
Differential equation
An equation involving the derivatives of an unknown function (introduced fully in Section 9, but relevant here through MVT corollaries).
Corollary (zero derivative implies constant)
If f'(x) = 0 for all x in (a, b), then f is constant on that interval. A direct consequence of the MVT.
Two conditions can prevent a function from having an absolute max or min on an interval:
Discontinuity on the interval. Consider f(x) = 1/x on [-2, 2]. The function blows up near x = 0, so it never attains a finite minimum. The EVT does not apply because f is not continuous on the entire closed interval.
Open endpoints. Consider f(x) = x on (0, 2]. The function gets arbitrarily close to 0 but never reaches it (the endpoint 0 is excluded), so there is no absolute minimum.
The lesson: continuity and a closed interval are both required for the EVT to guarantee extrema exist.
If f is continuous on [a, b], then f attains an absolute maximum value and an absolute minimum value somewhere in [a, b].
The theorem tells you extrema exist; it does not tell you where they are. For that, you need the Closed Interval Method.
A critical number c is a point in the domain of f where f'(c) = 0 or f'(c) does not exist. These are the only candidates for interior extrema.
Example: for f(x) = 2x^3 - 3x^2 - 12x + 5, the derivative is f'(x) = 6x^2 - 6x - 12 = 6(x - 2)(x + 1). Setting f'(x) = 0 gives critical numbers x = 2 and x = -1.
To find the absolute maximum and minimum of a continuous function f on a closed interval [a, b]:
Find the values of f at all critical numbers of f in (a, b).
Find the values of f at the endpoints a and b.
The largest value from steps 1 and 2 is the absolute maximum; the smallest is the absolute minimum.
Finding absolute extrema of f(x) = x^3 - 12x + 1 on [1, 4]
f'(x) = 3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2). Critical numbers: x = 2 and x = -2. Only x = 2 lies in (1, 4). Evaluate: f(1) = 1 - 12 + 1 = -10, f(2) = 8 - 24 + 1 = -15, f(4) = 64 - 48 + 1 = 17. Absolute max = 17 at x = 4. Absolute min = -15 at x = 2.
Finding absolute extrema of g(t) = t - 3t^(2/3) on [-1, 27]
g'(t) = 1 - 2t^(-1/3) = 1 - 2/(t^(1/3)). Setting g'(t) = 0 gives t^(1/3) = 2, so t = 8. Also, g'(t) is undefined at t = 0 (which is in the domain), so t = 0 is a critical number too. Evaluate: g(-1) = -1 - 3 = -4, g(0) = 0, g(8) = 8 - 12 = -4, g(27) = 27 - 27 = 0. Absolute max = 0 at t = 0 and t = 27. Absolute min = -4 at t = -1 and t = 8.
Finding critical numbers of f(t) = 7t^(2/3) - 9t^(5/3)
f'(t) = (14/3)t^(-1/3) - 15t^(2/3) = t^(-1/3)[(14/3) - 15t]. Setting the bracket to zero: t = 14/45. Also, f'(t) is undefined at t = 0. Critical numbers: t = 0 and t = 14/45.
Absolute extrema of g(s) = 1/(s - 5) on [-5, -4]
g'(s) = -1/(s - 5)^2, which is never zero, so there are no critical numbers in [-5, -4]. Check endpoints: g(-5) = 1/(-10) = -1/10, g(-4) = 1/(-9) = -1/9. Since -1/10 < -1/9, the absolute min is -1/10 at s = -5 and the absolute max is -1/9 at s = -4.
Rolle's Theorem is a special case of the MVT. Its three hypotheses are:
f is continuous on [a, b].
f is differentiable on (a, b).
f(a) = f(b).
Conclusion: there exists at least one c in (a, b) with f'(c) = 0. Geometrically, if the function starts and ends at the same height, the tangent line must be horizontal somewhere in between.
Hypotheses:
f is continuous on [a, b].
f is differentiable on (a, b).
Conclusion: there exists a c in (a, b) such that
f'(c) = \frac{f(b) - f(a)}{b - a}Geometrically, the tangent line at x = c is parallel to the secant line joining (a, f(a)) and (b, f(b)). The right-hand side is the average rate of change; the left-hand side is the instantaneous rate of change. The MVT says they must match somewhere.
Rolle's Theorem is the MVT with f(a) = f(b), which makes the right-hand side zero.
Before using the MVT, verify both hypotheses:
Continuity on the closed interval [a, b]. Watch for division by zero, square roots of negatives, or piecewise jumps.
Differentiability on the open interval (a, b). Watch for corners (absolute-value functions), cusps (fractional-exponent functions like x^(2/3) at 0), and vertical tangents.
If either hypothesis fails, the MVT does not apply and you cannot draw its conclusion.
f(x) = x^2 + 3x + 5 on [0, 1]
Polynomials are continuous and differentiable everywhere, so the MVT applies. Average rate of change: (f(1) - f(0))/(1 - 0) = (9 - 5)/1 = 4. f'(x) = 2x + 3. Set 2c + 3 = 4, so c = 1/2. This c lies in (0, 1), confirming the MVT.
f(x) = 1/x on [1, 3]
f is continuous and differentiable on [1, 3] (no zero in this interval). Average rate of change: (1/3 - 1)/(3 - 1) = (-2/3)/2 = -1/3. f'(x) = -1/x^2. Set -1/c^2 = -1/3, so c^2 = 3, giving c = sqrt(3) (approximately 1.73), which lies in (1, 3).
f(x) = |x| on [-2, 2]
f is continuous on [-2, 2], but f is not differentiable at x = 0 (corner point). Since differentiability on (-2, 2) fails, the MVT does not apply. Notice the average rate of change is (2 - 2)/(2 - (-2)) = 0, yet there is no c where f'(c) = 0 because f'(x) = -1 for x < 0 and f'(x) = 1 for x > 0.
f(x) = cube root of x on [0, 1]
f is continuous on [0, 1]. f'(x) = 1/(3x^(2/3)), which is undefined at x = 0, but the MVT requires differentiability only on the open interval (0, 1), and f' is defined there. Average rate of change: (1 - 0)/(1 - 0) = 1. Set 1/(3c^(2/3)) = 1, so c^(2/3) = 1/3, giving c = (1/3)^(3/2) = 1/(3*sqrt(3)), which is approximately 0.19 and lies in (0, 1).
Corollary 1: zero derivative means constant. If f'(x) = 0 for all x in (a, b), then f(x) is constant on (a, b). The proof uses the MVT: for any two points in the interval, the difference in function values equals f'(c) times the distance, which is zero.
Corollary 2: equal derivatives means functions differ by a constant. If f'(x) = g'(x) for all x in (a, b), then f(x) = g(x) + C for some constant C. This is the theoretical basis for the "+C" that appears in every antiderivative.
The MVT can show that an equation has at most one solution. Example: show 3x + 2cos(x) + 5 = 0 has at most one solution. Assume by contradiction there are two solutions, a and b. Then f(x) = 3x + 2cos(x) + 5 satisfies f(a) = f(b) = 0, so by Rolle's Theorem there exists c with f'(c) = 0. But f'(x) = 3 - 2sin(x), and since -1 <= sin(x) <= 1, we get 1 <= f'(x) <= 5. So f'(c) is never zero, a contradiction.
Closed Interval Method (finding absolute extrema on [a, b])
Compute f'(x) and find all c in (a, b) where f'(c) = 0 or f'(c) DNE.
Evaluate f at each critical number and at the endpoints a and b.
Largest value = absolute max; smallest = absolute min.
Rolle's Theorem
Continuous on [a, b], differentiable on (a, b), f(a) = f(b) implies there exists c in (a, b) with f'(c) = 0.
Mean Value Theorem
f'(c) = \frac{f(b) - f(a)}{b - a}for some c in (a, b), provided f is continuous on [a, b] and differentiable on (a, b).
Zero-derivative corollary
f'(x) = 0 on (a, b) implies f is constant on (a, b).
Equal-derivative corollary
f'(x) = g'(x) on (a, b) implies f(x) = g(x) + C for some constant C.
Optimisation in engineering and economics relies on finding maxima and minima. Sections 1 and 2 supply the theoretical foundation: the EVT tells you an optimal value exists (for continuous functions on closed intervals), the Closed Interval Method finds it, and the MVT provides the link between average and instantaneous rates. For instance, if a car travels 100 miles in 2 hours, the MVT guarantees the speedometer read exactly 50 mph at some moment during the trip, regardless of how speed varied.
Students often think the EVT applies to any interval. It does not. The interval must be closed [a, b] and the function must be continuous on the entire interval. An open interval or a discontinuity can ruin it.
Students sometimes assume that a critical number is automatically a local max or min. A critical number is only a candidate. The function f(x) = x^3 has a critical number at x = 0 (since f'(0) = 0), but x = 0 is neither a local max nor a local min.
Students often confuse the hypotheses of Rolle's Theorem and the MVT. Rolle's Theorem requires the extra condition f(a) = f(b). Without it, you can only conclude the MVT, not that f'(c) = 0.
A common error is applying the MVT when differentiability fails. The function f(x) = |x| on [-2, 2] is continuous but not differentiable at x = 0. The MVT cannot be applied, and indeed its conclusion fails: the average slope is 0 but f' is never 0.
⚠️ The Closed Interval Method is tested repeatedly. Know the three steps cold: find critical numbers in (a, b), evaluate f at those and at endpoints, compare.
⚠️ Exam questions often ask you to verify the hypotheses of the MVT before finding c. Do not skip this step. State continuity and differentiability explicitly.
⚠️ Be ready for functions where f'(c) does not exist (fractional exponents, absolute values). These create critical numbers that students frequently miss.
⚠️ Proof-by-contradiction problems using the MVT appear on exams. The structure: assume two solutions exist, apply Rolle's Theorem, show f' can never be zero, contradiction.
True or False: A continuous function on an open interval (a, b) always has an absolute maximum. Answer: False. The EVT requires a closed interval.
Fill in the blank: A critical number c satisfies either f'(c) = ____ or f'(c) ____. Answer: 0; does not exist.
True or False: The Mean Value Theorem tells you the exact value of c. Answer: False. It guarantees c exists but does not specify its value.
True or False: Rolle's Theorem is a special case of the MVT where f(a) = f(b). Answer: True.
Fill in the blank: If f'(x) = 0 for all x in (a, b), then f is ____ on (a, b). Answer: constant.
Q: Find the absolute maximum and minimum of f(x) = x^4 - 2x^2 + 1 on [-2, 1].
A: f'(x) = 4x^3 - 4x = 4x(x^2 - 1) = 4x(x - 1)(x + 1). Critical numbers in (-2, 1): x = -1 and x = 0. (x = 1 is an endpoint.) Evaluate: f(-2) = 16 - 8 + 1 = 9, f(-1) = 1 - 2 + 1 = 0, f(0) = 1, f(1) = 0. Absolute max = 9 at x = -2. Absolute min = 0 at x = -1 and x = 1.
Q: Does the MVT apply to f(x) = sqrt(x) on [0, 4]? If so, find c.
A: f is continuous on [0, 4] and differentiable on (0, 4) (f'(x) = 1/(2sqrt(x)) is defined for x > 0). Average rate of change: (2 - 0)/4 = 1/2. Set 1/(2sqrt(c)) = 1/2, so sqrt(c) = 1, giving c = 1, which is in (0, 4).
Q: Use the MVT to show that sin(x) < x for all x > 0.
A: Let f(t) = t - sin(t). Then f(0) = 0 and f'(t) = 1 - cos(t) >= 0 for all t, with equality only at isolated points. By the MVT, for any x > 0 there exists c in (0, x) with f(x) - f(0) = f'(c)*x > 0 (since f'(c) > 0 for almost all c). So f(x) > 0, meaning x > sin(x).
Q: Verify Rolle's Theorem for f(x) = x^2 - 4x + 3 on [1, 3].
A: f is a polynomial (continuous, differentiable everywhere). f(1) = 1 - 4 + 3 = 0, f(3) = 9 - 12 + 3 = 0, so f(1) = f(3). By Rolle's Theorem, there exists c in (1, 3) with f'(c) = 0. f'(x) = 2x - 4. Set 2c - 4 = 0, so c = 2, which is in (1, 3).
The critical-number machinery from Section 1 reappears in Section 3 (the First and Second Derivative Tests classify critical points as local max, local min, or neither) and in Section 7 (Optimisation Problems, where you set up a real-world quantity to maximise or minimise). The MVT's corollaries connect directly to Section 9 (Antiderivatives), because the fact that equal derivatives imply functions differ by a constant is what justifies the "+C" in indefinite integrals. In multivariable calculus (MTH 234), the Second Derivative Test extends to functions of two variables.
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