Difficulty: Intermediate | Prerequisites: Basic differentiation (power rule, product rule, chain rule), setting equations equal to zero and solving.
Optimization is the process of finding and classifying the highest or lowest points on a function's graph. You find candidates (called critical values) by setting the first derivative equal to zero or identifying where it does not exist, then use the First Derivative Test to check whether each candidate is a relative maximum, a relative minimum, or neither. This technique applies directly to business and science problems where you need to maximise revenue, minimise cost, or find peak output.
Tags: optimization, critical value, critical point, relative extrema, first derivative test, sign chart, local max, local min
Optimization
The process of finding and interpreting the highest or lowest points (extrema) on a graph.
Think of it as: figuring out where a function hits its peak or its valley, and what that means in context.
Extrema
A general term for maximum and minimum points on a graph, whether local or absolute.
In simple terms, these are the "hilltops" and "valley floors" of a curve.
Relative (local) maximum
A point where the function value is higher than at all nearby points in its immediate neighbourhood. The function rises to this point, then falls away from it.
Think of it as: the top of a hill. You are higher than everything around you, but there might be a taller mountain further off.
Relative (local) minimum
A point where the function value is lower than at all nearby points in its immediate neighbourhood. The function falls to this point, then rises away from it.
Think of it as: the bottom of a dip. You are lower than everything around you, but there might be a deeper valley elsewhere.
Critical value
An x-value in the domain of f where f'(x) = 0 or where f'(x) does not exist.
In simple terms, these are the x-coordinates where the graph might have a peak, valley, or corner.
Critical point
The point (x, f(x)) on the graph corresponding to a critical value. It is the full coordinate pair, not just the x-value.
Think of it as: the critical value gives you the horizontal position; the critical point gives you the complete location on the curve.
First Derivative Test
A method for classifying critical points by checking whether the sign of f'(x) changes from positive to negative, negative to positive, or stays the same as you move through the critical value.
In simple terms, you plug test points into the derivative on either side of the critical value and see whether the function was going up or going down.
This topic sits in the "Applications of Derivatives" unit. You have already learned how to take derivatives; now you are using them to answer practical questions: where does a function reach its highest or lowest value, and what is that value? This matters because most real-world optimisation problems (maximise profit, minimise waste, find peak demand) come down to finding and classifying extrema. You need solid differentiation skills (power rule, product rule, chain rule) before tackling this material.
For a continuous function y = f(x), possible maximum or minimum points exist only at x-values where:
The first derivative equals zero: f'(x) = 0 (the tangent line is flat)
The first derivative does not exist (the graph has a corner, cusp, or vertical tangent)
These x-values are called critical values. The corresponding points (x, f(x)) on the graph are called critical points.
A key nuance: every relative extremum occurs at a critical value, but not every critical value produces a relative extremum. The converse holds, not the biconditional.
Once you have the critical values, classify each one by checking the sign of f'(x) on either side:
Positive to negative (f' changes from + to -): the function rises then falls, so the critical point is a relative maximum
Negative to positive (f' changes from - to +): the function falls then rises, so the critical point is a relative minimum
No sign change (f' stays + on both sides, or stays - on both sides): the critical point is neither a max nor a min
To perform the test, pick a convenient test point in each subinterval created by the critical values, plug it into f'(x), and note whether the result is positive or negative. You do not need the exact value, only the sign.
Let f(x) = x³ - 3x² + 2x + 4. Find all relative extrema and classify them.
Step 1: Find the derivative.
f'(x) = 3x² - 6x + 2
Step 2: Set the derivative equal to zero and solve.
Using the quadratic formula on 3x² - 6x + 2 = 0:
x = (6 ± √(36 - 24)) / 6 = (6 ± √12) / 6
x ≈ 0.423 and x ≈ 1.577
Step 3: Find the critical points.
Plug each critical value back into the original function:
f(0.423) ≈ 4.385, giving the critical point (0.423, 4.385)
f(1.577) ≈ 3.615, giving the critical point (1.577, 3.615)
Step 4: Build a sign chart.
The two critical values split the number line into three subintervals:
Subinterval (-∞, 0.423): test point x = 0, f'(0) = 2 (positive)
Subinterval (0.423, 1.577): test point x = 1, f'(1) = 3 - 6 + 2 = -1 (negative)
Subinterval (1.577, ∞): test point x = 2, f'(2) = 12 - 12 + 2 = 2 (positive)
Step 5: Classify.
At x ≈ 0.423, f' changes from + to -, so (0.423, 4.385) is a relative maximum
At x ≈ 1.577, f' changes from - to +, so (1.577, 3.615) is a relative minimum
Quadratic formula (used to solve f'(x) = 0 when the derivative is quadratic):
x = (-b ± √(b² - 4ac)) / 2a
Product rule (used when differentiating a product of two functions):
If y = u(x) · v(x), then y' = u'(x) · v(x) + u(x) · v'(x)
Derivative of e^(kx):
d/dx [e^(kx)] = k · e^(kx)
Sign chart template:
Find all critical values (where f'(x) = 0 or DNE)
Place them on a number line, dividing it into subintervals
Choose one test point per subinterval
Evaluate f'(test point) and record the sign (+ or -)
Read the sign changes across each critical value to classify
A product launches online and the number of daily orders is modelled by S(t) = 200t · e^(-0.006t), where t is the number of days since launch. The question: on which day do orders peak, and how many orders come in on that day?
This is a classic "find the maximum" optimisation problem using the product rule.
Derivative (via product rule):
S'(t) = 200e^(-0.006t) + 200t · (-0.006)e^(-0.006t)
S'(t) = 200e^(-0.006t)(1 - 0.006t)
Setting S'(t) = 0:
Since 200e^(-0.006t) is never zero (exponentials are always positive), the only solution comes from:
1 - 0.006t = 0, so t = 1/0.006 ≈ 166.667
Rounding appropriately, the critical value is t ≈ 167 days.
Sign chart:
On (0, 167): S'(t) > 0, so the graph is increasing (orders are growing)
On (167, ∞): S'(t) < 0, so the graph is decreasing (orders are falling)
The sign changes from positive to negative at t ≈ 167, confirming this is a relative maximum.
Number of orders on that day:
S(167) = 200(167)e^(-0.006 × 167) ≈ 200(167)e^(-1.002) ≈ 200(167)(0.3673) ≈ 12,267 orders
On the 167th day after launch, orders peak at roughly 12,267. After that, daily order volume declines.
This sort of model appears whenever a product or campaign sees an initial surge followed by a gradual tail-off. The exponential decay factor e^(-0.006t) eventually overwhelms the linear growth factor 200t.
Students often assume that every critical value produces a maximum or minimum. It does not. If the derivative does not change sign at a critical value, that point is neither a max nor a min (for example, x = 0 for f(x) = x³).
Students sometimes confuse "critical value" with "critical point." The critical value is the x-coordinate alone. The critical point is the full ordered pair (x, f(x)).
When using the First Derivative Test, students sometimes plug test points into the original function rather than the derivative. The sign chart is built from f'(x), not f(x).
Students occasionally forget that critical values also arise where f'(x) does not exist, not only where f'(x) = 0. Corners and cusps on the graph are critical values too.
⚠️ You will almost certainly be asked to find critical values and classify them using the First Derivative Test. Know the procedure cold: differentiate, set equal to zero, solve, build a sign chart, classify.
⚠️ Applied problems (revenue, orders, cost) are a favourite exam format. Expect to build a model from a word problem, differentiate it, and interpret the result in context ("On day 167, orders peak at 12,267").
⚠️ Be precise with rounding. The source material specifies three decimal places for coordinates. Follow whatever precision the problem states.
⚠️ The distinction between relative and absolute extrema is a common exam question. Relative extrema are local (highest or lowest in their neighbourhood). Absolute extrema are global (highest or lowest on the entire domain or interval). Part C of this lesson (covered in the companion study notes) introduces the Extreme Value Theorem for absolute extrema.
True or false: every critical value of a continuous function corresponds to a relative maximum or minimum.
Answer: False. A critical value can be neither if the derivative does not change sign.
Fill in the blank: a critical value is an x-value where f'(x) = ______ or where f'(x) ____________.
Answer: 0; does not exist.
True or false: if f'(x) changes from negative to positive at a critical value, that critical point is a relative maximum.
Answer: False. Negative to positive indicates a relative minimum (the function falls, then rises).
Fill in the blank: the critical points of f(x) = x³ - 3x² + 2x + 4 are approximately (, 4.385) and (, 3.615).
Answer: 0.423; 1.577.
True or false: to build a sign chart, you plug test points into the original function f(x).
Answer: False. You plug test points into the derivative f'(x).
Q: Given f(x) = x³ - 3x² + 2x + 4, find f'(x) and identify the critical values.
A: f'(x) = 3x² - 6x + 2. Setting 3x² - 6x + 2 = 0 and using the quadratic formula gives x ≈ 0.423 and x ≈ 1.577.
Q: For the function above, use the First Derivative Test to classify the critical point at x ≈ 0.423.
A: Test x = 0 (left): f'(0) = 2 > 0. Test x = 1 (right): f'(1) = -1 < 0. The derivative changes from positive to negative, so (0.423, 4.385) is a relative maximum.
Q: The daily orders for a product are modelled by S(t) = 200te^(-0.006t). Find S'(t) using the product rule.
A: S'(t) = 200e^(-0.006t) + 200t(-0.006)e^(-0.006t) = 200e^(-0.006t)(1 - 0.006t).
Q: At what value of t is the number of daily orders maximised for S(t) = 200te^(-0.006t)?
A: Set 1 - 0.006t = 0, giving t ≈ 166.667, which rounds to approximately day 167.
Q: Explain why the exponential factor e^(-0.006t) in S'(t) = 200e^(-0.006t)(1 - 0.006t) does not contribute any critical values.
A: The exponential function e^(-0.006t) is always positive for all real t. Since it is never zero and never undefined, it cannot make the product equal zero. Only the factor (1 - 0.006t) can equal zero.
This connects to the Second Derivative Test (covered in the next MAT210 lesson), which provides an alternative way to classify critical points using concavity rather than a sign chart. If f''(c) > 0 at a critical value c, the point is a minimum; if f''(c) < 0, it is a maximum.
It also connects to curve sketching, where you combine information about critical values, intervals of increase/decrease, concavity, and inflection points to draw a complete graph of a function.
The Extreme Value Theorem, covered in the companion study notes, extends these ideas to closed intervals, guaranteeing the existence of absolute maxima and minima.
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