Calculus 12 Unit 4 Notes: Extrema, Curve
Sketching & Optimization
A Student-Friendly Guide for Tests, Quizzes, and Final Review
These notes cover the main skills students usually need for a Calculus 12 extrema unit: increasing/
decreasing intervals, local and absolute extrema, concavity, inflection points, curve sketching, and
optimization word problems.
1. Big Picture: What Is This Unit About?
This unit is about using derivatives to understand the shape and behaviour of functions.
The first derivative tells us about:
• where a function is increasing or decreasing
• where local maximums and minimums may happen
• the slope of the tangent line
The second derivative tells us about:
• whether the graph is concave up or concave down
• where inflection points may happen
• whether a critical point is a local maximum or local minimum using the second derivative test
In simple terms:
Derivative Main Job
f'(x) Increasing/decreasing and possible max/min points
f''(x) Concavity and possible inflection points
2. Critical Numbers
Definition
A critical number is a value of x in the domain of f(x) where:
1. f'(x) = 0 , or
1
, 2. f'(x) does not exist.
Critical numbers are important because local maximums and local minimums can only happen at critical
numbers or endpoints.
How to Find Critical Numbers
Step 1: Find f'(x) .
Step 2: Solve f'(x) = 0 .
Step 3: Check where f'(x) does not exist.
Step 4: Make sure the x -values are in the domain of the original function.
Example 1: Polynomial
Find the critical numbers of:
f(x) = x^3 - 6x^2 + 9x + 1
Solution
Differentiate:
f'(x) = 3x^2 - 12x + 9
Set the derivative equal to zero:
3x^2 - 12x + 9 = 0
Divide by 3:
x^2 - 4x + 3 = 0
Factor:
(x - 1)(x - 3) = 0
So:
x = 1 and x = 3
2
, Answer
The critical numbers are:
x = 1, 3
Example 2: Rational Function
Find the critical numbers of:
f(x) = \frac{x}{x^2 + 1}
Solution
Use the quotient rule:
f'(x) = \frac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2}
Simplify:
f'(x) = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2}
f'(x) = \frac{1 - x^2}{(x^2 + 1)^2}
Set the numerator equal to zero:
1 - x^2 = 0
x^2 = 1
x = \pm 1
The denominator is never zero because x^2 + 1 is always positive.
Answer
The critical numbers are:
x = -1, 1
3
Sketching & Optimization
A Student-Friendly Guide for Tests, Quizzes, and Final Review
These notes cover the main skills students usually need for a Calculus 12 extrema unit: increasing/
decreasing intervals, local and absolute extrema, concavity, inflection points, curve sketching, and
optimization word problems.
1. Big Picture: What Is This Unit About?
This unit is about using derivatives to understand the shape and behaviour of functions.
The first derivative tells us about:
• where a function is increasing or decreasing
• where local maximums and minimums may happen
• the slope of the tangent line
The second derivative tells us about:
• whether the graph is concave up or concave down
• where inflection points may happen
• whether a critical point is a local maximum or local minimum using the second derivative test
In simple terms:
Derivative Main Job
f'(x) Increasing/decreasing and possible max/min points
f''(x) Concavity and possible inflection points
2. Critical Numbers
Definition
A critical number is a value of x in the domain of f(x) where:
1. f'(x) = 0 , or
1
, 2. f'(x) does not exist.
Critical numbers are important because local maximums and local minimums can only happen at critical
numbers or endpoints.
How to Find Critical Numbers
Step 1: Find f'(x) .
Step 2: Solve f'(x) = 0 .
Step 3: Check where f'(x) does not exist.
Step 4: Make sure the x -values are in the domain of the original function.
Example 1: Polynomial
Find the critical numbers of:
f(x) = x^3 - 6x^2 + 9x + 1
Solution
Differentiate:
f'(x) = 3x^2 - 12x + 9
Set the derivative equal to zero:
3x^2 - 12x + 9 = 0
Divide by 3:
x^2 - 4x + 3 = 0
Factor:
(x - 1)(x - 3) = 0
So:
x = 1 and x = 3
2
, Answer
The critical numbers are:
x = 1, 3
Example 2: Rational Function
Find the critical numbers of:
f(x) = \frac{x}{x^2 + 1}
Solution
Use the quotient rule:
f'(x) = \frac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2}
Simplify:
f'(x) = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2}
f'(x) = \frac{1 - x^2}{(x^2 + 1)^2}
Set the numerator equal to zero:
1 - x^2 = 0
x^2 = 1
x = \pm 1
The denominator is never zero because x^2 + 1 is always positive.
Answer
The critical numbers are:
x = -1, 1
3
