Difficulty: Intermediate | Prerequisites: Basic derivative rules, chain rule, implicit differentiation
This covers the "what can derivatives tell us?" portion of Calc I: whether a function has a guaranteed maximum or minimum, how fast related quantities change together, and how to use a tangent line to estimate values near a known point. These are the applied problems that show up on the second half of most Calc I exams. If you are comfortable taking derivatives (chain rule, implicit differentiation), you have the mechanical skills. The challenge here is setting up the problem correctly before differentiating.
This material tests three applications of derivatives: finding absolute extrema on closed intervals (Extreme Value Theorem + critical points), related rates (implicit differentiation with respect to time), and linear approximation (tangent-line estimates). The setup is harder than the calculus in every case.
Absolute maximum / absolute minimum
The largest or smallest value a function takes over an entire domain (or a specified interval). In simple terms, the highest peak and the lowest valley on the whole stretch you are looking at.
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 on [a, b]. Think of it as: a continuous function on a closed interval cannot "escape" to infinity, so it must have a top and a bottom.
Critical point / critical number
A value x = c in the domain of f where f'(c) = 0 or f'(c) does not exist. These are the only interior candidates for local (and therefore absolute) extrema.
Related rates
A class of problems where two or more quantities change with time, connected by an equation. You differentiate the equation implicitly with respect to t, then solve for the unknown rate. In simple terms, if you know how fast one thing is changing, you can figure out how fast a related thing is changing.
Linear approximation
Using the tangent line at x = a to estimate f(x) for x near a. The formula is L(x) = f(a) + f'(a)(x − a). Think of it as replacing a curve with the straight line that just touches it at one point.
The exam tests this in two ways: a conceptual question (which functions have an absolute max over their domain?) and a computational question (find the absolute max and min on a closed interval).
Conceptual question (from the exam)
For which functions is there an absolute maximum over the given domain?
f(x) = 3 − x⁴ over (−∞, ∞): Yes. This is a downward-opening quartic. It reaches its peak at x = 0, where f(0) = 3. Even though the domain is all of R, the function has a global maximum because it goes to −∞ in both directions.
f(x) = ln(x) over (0, ∞): No. ln(x) increases without bound as x → ∞. There is no ceiling.
f(x) = tan(x) over (−π/2, π/2): No. tan(x) → +∞ as x → π/2⁻. The open interval means the function never reaches the endpoints where it blows up, but it also never levels off.
f(x) = 1/x over [1, ∞): Yes. The maximum occurs at the left endpoint: f(1) = 1. The function decreases from there toward 0 as x → ∞, so 1 is the absolute maximum value.
The lesson: the Extreme Value Theorem guarantees extrema on closed, bounded intervals with continuous functions. But extrema can still exist outside that setting, you just have to check by understanding the function's behaviour.
This is a standard algorithm. The exam tests it on f(x) = x³ − 3x² − 9x + 5 over [−4, 4].
Step 1: Justify that extrema exist
f(x) is a polynomial, so it is continuous everywhere. A continuous function on a closed interval [a, b] must attain an absolute maximum and minimum by the Extreme Value Theorem.
Step 2: Find critical points
Set f'(x) = 0. Here f'(x) = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1).
Critical points: x = 3 and x = −1. Both lie inside [−4, 4], so both are candidates.
Step 3: Evaluate f at critical points and endpoints
f(−4) = (−4)³ − 3(−4)² − 9(−4) + 5 = −64 − 48 + 36 + 5 = −71
f(−1) = (−1)³ − 3(−1)² − 9(−1) + 5 = −1 − 3 + 9 + 5 = 10
f(3) = 27 − 27 − 27 + 5 = −22
f(4) = 64 − 48 − 36 + 5 = −15
Step 4: Compare
The largest value is 10 (at x = −1), so the absolute maximum value is 10.
The smallest value is −71 (at x = −4), so the absolute minimum value is −71.
The method works every time on a closed interval: find the critical points, evaluate f at those plus the endpoints, pick the biggest and smallest.
Related rates problems give you a geometric or physical setup, tell you how fast one quantity is changing, and ask for the rate of change of another quantity. The calculus is implicit differentiation with respect to time t. The hard part is the setup.
Worked example from the exam: the sliding ladder
A 13 ft ladder leans against a wall. The bottom is 5 ft from the wall. The top slides down at 2 ft/min. Find the rate at which the angle between the ladder and the floor is changing.
Step 1: Draw and label
Let x = distance from wall to base, y = height of top on wall, θ = angle between ladder and floor. The ladder is the hypotenuse of a right triangle, so x² + y² = 13² = 169.
Step 2: Identify knowns and unknowns
x = 5 (given)
y = √(169 − 25) = √144 = 12
dy/dt = −2 ft/min (the top slides down, so y is decreasing)
Want: dθ/dt
Step 3: Choose an equation relating θ to known quantities
Use sin θ = y/13 (the ladder length is constant, and y is the side opposite θ).
Step 4: Differentiate with respect to t
cos θ · dθ/dt = (1/13) · dy/dt
Step 5: Solve for dθ/dt
cos θ = x/13 = 5/13 (from the triangle).
(5/13) · dθ/dt = (1/13)(−2)
dθ/dt = (−2/13) / (5/13) = −2/5
dθ/dt = −2/5 rad/min
The negative sign makes sense: the angle is getting smaller as the ladder slides down.
General related rates strategy
Draw a picture and assign variables to changing quantities
Write down what you know and what you want
Find an equation relating the variables (often Pythagorean theorem, similar triangles, trig, or a volume/area formula)
Differentiate both sides with respect to t
Substitute known values and solve for the unknown rate
Include units and interpret the sign
Linear approximation replaces a function with its tangent line near a known point. It is useful for estimating values that are close to something you can compute exactly.
The formula
L(x) = f(a) + f'(a)(x − a)
This is the tangent line to y = f(x) at x = a, rewritten as an approximation.
Worked example from the exam: f(x) = ln(x³ − 7) at a = 2
Step 1: Compute f(a).
f(2) = ln(2³ − 7) = ln(8 − 7) = ln(1) = 0
Step 2: Compute f'(x), then f'(a).
f'(x) = 3x² / (x³ − 7) (chain rule: derivative of ln(u) is 1/u times u')
f'(2) = 3(4) / (8 − 7) = 12/1 = 12
Step 3: Write the linear approximation.
L(x) = 0 + 12(x − 2) = 12(x − 2)
So ln(x³ − 7) ≈ 12(x − 2) for x near 2.
Using the approximation
To estimate ln((2.01)³ − 7): plug x = 2.01 into the linear approximation.
L(2.01) = 12(2.01 − 2) = 12(0.01) = 0.12
So ln((2.01)³ − 7) ≈ 0.12.
The approximation is good because 2.01 is very close to 2. The further x moves from a, the worse the estimate becomes.
Real-world application
Engineers use linear approximation constantly when a full computation is too complex or when they need a quick estimate. Any time a measurement has a small error Δx, the effect on a computed quantity is approximately f'(a) · Δx. This is the basis of error propagation in experimental science.
Extreme Value Theorem conditions
f must be continuous on [a, b]
The interval must be closed and bounded
Then f has both an absolute max and an absolute min on [a, b]
Closed-interval method
Find all critical points of f in (a, b) where f'(x) = 0 or f'(x) DNE
Evaluate f at each critical point and at the endpoints a and b
The largest value is the absolute max; the smallest is the absolute min
Linear approximation
L(x) = f(a) + f'(a)(x − a)
Common geometric relationships for related rates
Pythagorean theorem: x² + y² = L² (right triangle, ladder, distance)
Trigonometric: sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse
Volume of a cone: V = (1/3)πr²h
Volume of a sphere: V = (4/3)πr³
Area of a circle: A = πr²
Students assume that the Extreme Value Theorem applies to open intervals. It does not. The theorem requires a closed interval [a, b]. On an open interval, a continuous function may fail to reach a maximum or minimum.
Students forget to check the endpoints when finding absolute extrema on a closed interval. The absolute max or min often occurs at an endpoint, not at a critical point.
In related rates, students substitute numerical values before differentiating. You must differentiate first (keeping variables), then substitute the known values. Plugging in numbers too early kills the rates.
Students confuse "rate of change of the angle" with "rate of change of a side." Read the problem carefully to identify which rate is given and which is asked for.
In linear approximation, students sometimes forget that the approximation is only good near x = a. Using L(x) for x far from a gives a poor estimate.
⚠️ The absolute extrema problem on a closed interval is almost guaranteed. It is a straightforward algorithm, so there is no reason to lose marks here. Memorise the three steps.
⚠️ The related rates problem requires you to show all work. Points are given for the setup (diagram, variables, equation) as much as for the differentiation. A correct final answer with no supporting work earns zero.
⚠️ Linear approximation problems typically have two parts: derive the approximation formula, then use it to estimate a specific value. The second part is easy if the first is correct, so double-check your derivative.
⚠️ The conceptual "which functions have an absolute max" question tests understanding of the Extreme Value Theorem and function behaviour, with no computation required. Sketching the function helps.
⚠️ Units and signs matter in related rates. The exam specifically asks you to include both. A negative rate means the quantity is decreasing.
True or False: A continuous function on an open interval (a, b) must have an absolute maximum. (False: the EVT requires a closed interval.)
Fill in the blank: The linear approximation of f(x) near x = a is L(x) = f(a) + ______ · (x − a). (f'(a).)
True or False: In a related rates problem, you should substitute known values before differentiating. (False: differentiate first, then substitute.)
True or False: f(x) = 3 − x⁴ has an absolute maximum over (−∞, ∞). (True: the max is 3 at x = 0.)
Fill in the blank: To find the absolute max of a continuous function on [a, b], evaluate f at ______ and at the ______. (Critical points; endpoints.)
Q: Find the absolute maximum and minimum of g(x) = 2x³ − 3x² − 12x + 1 on [−2, 3].
A: g'(x) = 6x² − 6x − 12 = 6(x² − x − 2) = 6(x − 2)(x + 1). Critical points: x = 2 and x = −1, both in [−2, 3]. Evaluate: g(−2) = −16 − 12 + 24 + 1 = −3; g(−1) = −2 − 3 + 12 + 1 = 8; g(2) = 16 − 12 − 24 + 1 = −19; g(3) = 54 − 27 − 36 + 1 = −8. Absolute max = 8 at x = −1. Absolute min = −19 at x = 2.
Q: A spherical balloon is being inflated at 100 cm³/s. How fast is the radius increasing when r = 5 cm?
A: V = (4/3)πr³, so dV/dt = 4πr² · dr/dt. Solve: dr/dt = (dV/dt) / (4πr²) = 100 / (4π · 25) = 100/(100π) = 1/π ≈ 0.318 cm/s.
Q: Find the linear approximation of f(x) = √x at a = 25, and use it to estimate √26.
A: f(25) = 5, f'(x) = 1/(2√x), so f'(25) = 1/10. L(x) = 5 + (1/10)(x − 25). At x = 26: L(26) = 5 + 1/10 = 5.1.
Q: Does f(x) = 1/(x − 1) have an absolute maximum on (0, 1)?
A: No. As x → 1⁻, f(x) → +∞. The function is unbounded above on this open interval.
Q: Why can the Extreme Value Theorem not be applied to f(x) = 1/x on (0, 2)?
A: Two reasons: the interval (0, 2) is open (not closed), and f is not defined (hence not continuous) at x = 0. The EVT requires f to be continuous on a closed, bounded interval.
Related rates problems are the applied side of implicit differentiation, which is covered in detail in the companion study notes on differentiation techniques. The Extreme Value Theorem connects forward to optimisation problems (finding the dimensions that maximise area or minimise cost), which typically appear on Exam 3. Linear approximation is the first term of a Taylor series, a major topic in Calc II.
absolute maximum, absolute minimum, Extreme Value Theorem, EVT, closed interval method, critical points, critical numbers, related rates, sliding ladder problem, implicit differentiation with respect to time, dθ/dt, linear approximation, tangent line approximation, linearisation, L(x) = f(a) + f'(a)(x − a), MA-UY 1024, Calc 1 NYU, Calculus II for Engineers, Exam 2 review, optimisation setup, error propagation