Skip to content Study Mondo Free study resources for students from Grade 4 through AP and test prep. 24 courses, 700+ topics.
Courses Features Company Stay Ahead in School Free weekly study tips, practice sets, and exam strategies. Join 10,000+ students.
ยฉ 2026 Study Mondo. Built for students.
APยฎ is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this website.
Critical Points and Extrema | Study Mondo
Topics / Applications of Derivatives / Critical Points and Extrema Critical Points and Extrema Finding maximum and minimum values of functions
๐ฏ โญ INTERACTIVE LESSON
Try the Interactive Version! Learn step-by-step with practice exercises built right in.
Start Interactive Lesson โ ๐ Critical Points and Extrema
What are Extrema?
Extrema are the maximum and minimum values of a function. They come in two types:
Types of Extrema
Absolute (Global) Extrema :
Absolute maximum : The highest point on the entire graph
Absolute minimum : The lowest point on the entire graph
Local (Relative) Extrema :
Local maximum : A peak - higher than nearby points
Local minimum : A valley - lower than nearby points
๐ก Key Idea : A function can have multiple local extrema, but at most one absolute max and one absolute min (on a given interval).
What are Critical Points?
A critical point occurs at if:
๐ Practice Problems
1 Problem 1easy โ Question:Find all critical points of f ( x ) = x 4 โ 4 x 3 + 3 f(x) = x^4 - 4x^3 + 3 f ( x ) = x 4 โ .
Explain using: ๐ Simple words ๐ Analogy ๐จ Visual desc. ๐ Example ๐ก Explain
๐ AP Calculus AB โ Exam Format Guideโฑ 3 hours 15 minutes ๐ 51 questions ๐ 4 sections
Section Format Questions Time Weight Calculator Multiple Choice (No Calculator) MCQ 30 60 min 33.3% ๐ซ Multiple Choice (Calculator) MCQ 15 45 min 16.7% โ
Free Response (Calculator) FRQ 2 30 min 16.7% โ
Free Response (No Calculator) FRQ 4 60 min 33.3% ๐ซ
๐ก Key Test-Day Tipsโ Show all work on FRQsโ Use proper notationโ Check unitsโ Manage your timeโ ๏ธ Common Mistakes: Critical Points and ExtremaAvoid these 4 frequent errors
1 Forgetting the constant of integration (+C) on indefinite integrals
โพ 2 Confusing the Power Rule with the Chain Rule
โพ 3 Not checking continuity before applying the Mean Value Theorem
โพ 4 Dropping negative signs when differentiating trig functions
โพ ๐ Real-World Applications: Critical Points and ExtremaSee how this math is used in the real world
โ๏ธ Optimizing Package Design
Engineering
โพ ๐ฅ Predicting Drug Dosage Decay
Medicine
โพ ๐ฌ Calculating Distance from Velocity
Physics
โพ ๐ฐ Revenue Optimization
Finance
โพ
๐ Worked Example: Related Rates โ Expanding CircleProblem: A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 2 2 2 cm/s. How fast is the area of the circle increasing when the radius is 10 10 10 cm?
1 Identify the known and unknown rates Click to reveal โ
2 Write the relationship between variables
3 Differentiate both sides with respect to time
๐งช Practice Lab Interactive practice problems for Critical Points and Extrema
โพ ๐ Related Topics in Applications of Derivativesโ Frequently Asked QuestionsWhat is Critical Points and Extrema?โพ Finding maximum and minimum values of functions
How can I study Critical Points and Extrema effectively?โพ Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Critical Points and Extrema study guide free?โพ Yes โ all study notes, flashcards, and practice problems for Critical Points and Extrema on Study Mondo are 100% free. No account is needed to access the content.
What course covers Critical Points and Extrema?โพ Critical Points and Extrema is part of the AP Calculus AB course on Study Mondo, specifically in the Applications of Derivatives section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Critical Points and Extrema?
๐ก Study Tipsโ Work through examples step-by-step โ Practice with flashcards daily โ Review common mistakes x = c x = c x = c
f โฒ ( c ) = 0 f'(c) = 0 f โฒ ( c ) = 0 (horizontal tangent), OR
f โฒ ( c ) f'(c) f โฒ ( c ) does not exist (sharp corner, vertical tangent, discontinuity)
โ ๏ธ Important : Not all critical points are extrema! But all local extrema (that occur in the interior of the domain) ARE critical points.
Fermat's Theorem Fermat's Theorem : If f f f has a local extremum at x = c x = c x = c , and f โฒ ( c ) f'(c) f โฒ ( c ) exists, then f โฒ ( c ) = 0 f'(c) = 0 f โฒ ( c ) = 0 .
Local maxima and minima can only occur where the derivative is zero or undefined
To find extrema, we must check critical points!
BUT : Not every critical point is an extremum. We need to test them!
Finding Critical Points
Step-by-Step Process Step 1 : Find f โฒ ( x ) f'(x) f โฒ ( x )
Step 2 : Solve f โฒ ( x ) = 0 f'(x) = 0 f โฒ ( x ) = 0
Step 3 : Find where f โฒ ( x ) f'(x) f โฒ ( x ) is undefined
Step 4 : Check that these x x x -values are in the domain of f f f
All these points are critical points!
Example 1: Basic Polynomial Find all critical points of f ( x ) = x 3 โ 3 x 2 โ 9 x + 5 f(x) = x^3 - 3x^2 - 9x + 5 f ( x ) = x 3 โ 3 x 2 โ 9 x + 5
Step 1: Find the derivative
f โฒ ( x ) = 3 x 2 โ 6 x โ 9 f'(x) = 3x^2 - 6x - 9 f โฒ ( x ) = 3 x 2 โ 6 x โ 9
Step 2: Set equal to zero
3 x 2 โ 6 x โ 9 = 0 3x^2 - 6x - 9 = 0 3 x 2 โ 6 x โ 9 = 0
3 ( x 2 โ 2 x โ 3 ) = 0 3(x^2 - 2x - 3) = 0 3 ( x 2 โ 2 x โ 3 ) = 0
3 ( x โ 3 ) ( x + 1 ) = 0 3(x - 3)(x + 1) = 0 3 ( x โ 3 ) ( x + 1 ) = 0
So x = 3 x = 3 x = 3 or x = โ 1 x = -1 x = โ 1
Step 3: Check if undefined
f โฒ ( x ) f'(x) f โฒ ( x ) is a polynomial, so it's defined everywhere โ
Answer : Critical points at x = โ 1 x = -1 x = โ 1 and x = 3 x = 3 x = 3
Example 2: Rational Function Find critical points of f ( x ) = x 2 x โ 2 f(x) = \frac{x^2}{x - 2} f ( x ) = x โ 2 x 2 โ
Step 1: Find derivative (Quotient Rule)
f โฒ ( x ) = ( 2 x ) ( x โ 2 ) โ ( x 2 ) ( 1 ) ( x โ 2 ) 2 f'(x) = \frac{(2x)(x-2) - (x^2)(1)}{(x-2)^2} f โฒ ( x ) = ( x โ 2 ) 2 ( 2 x ) ( x โ 2 ) โ ( x 2 ) ( 1 ) โ
= 2 x 2 โ 4 x โ x 2 ( x โ 2 ) 2 = \frac{2x^2 - 4x - x^2}{(x-2)^2} = ( x โ 2 ) 2 2 x 2 โ 4 x โ x 2 โ
= x 2 โ 4 x ( x โ 2 ) 2 = \frac{x^2 - 4x}{(x-2)^2} = ( x โ 2 ) 2 x 2 โ 4 x โ
= x ( x โ 4 ) ( x โ 2 ) 2 = \frac{x(x-4)}{(x-2)^2} = ( x โ 2 ) 2 x ( x โ 4 ) โ
Step 2: Set equal to zero
x ( x โ 4 ) ( x โ 2 ) 2 = 0 \frac{x(x-4)}{(x-2)^2} = 0 ( x โ 2 ) 2 x ( x โ 4 ) โ = 0
Numerator = 0: x ( x โ 4 ) = 0 x(x-4) = 0 x ( x โ 4 ) = 0
So x = 0 x = 0 x = 0 or x = 4 x = 4 x = 4
Step 3: Check if undefined
f โฒ ( x ) f'(x) f โฒ ( x ) is undefined when ( x โ 2 ) 2 = 0 (x-2)^2 = 0 ( x โ 2 ) 2 = 0 , i.e., when x = 2 x = 2 x = 2
But f ( 2 ) f(2) f ( 2 ) is also undefined (not in domain), so x = 2 x = 2 x = 2 is NOT a critical point.
Answer : Critical points at x = 0 x = 0 x = 0 and x = 4 x = 4 x = 4
The Extreme Value Theorem Extreme Value Theorem (EVT) : If f f f is continuous on a closed interval [ a , b ] [a, b] [ a , b ] , then f f f has both an absolute maximum and an absolute minimum on [ a , b ] [a, b] [ a , b ] .
Why This Matters On a closed interval, extrema can occur at:
Critical points in the interior ( a , b ) (a, b) ( a , b )
Endpoints x = a x = a x = a or x = b x = b x = b
So to find absolute extrema on [ a , b ] [a, b] [ a , b ] :
Find all critical points in ( a , b ) (a, b) ( a , b )
Evaluate f f f at critical points AND endpoints
The largest value is the absolute max
The smallest value is the absolute min
Finding Absolute Extrema on a Closed Interval
The Closed Interval Method Step 1 : Find all critical points in ( a , b ) (a, b) ( a , b )
Step 2 : Evaluate f f f at each critical point
Step 3 : Evaluate f f f at the endpoints a a a and b b b
Step 4 : Compare all values:
Largest value = absolute maximum
Smallest value = absolute minimum
Example: Closed Interval Method Find the absolute maximum and minimum of f ( x ) = x 3 โ 3 x + 1 f(x) = x^3 - 3x + 1 f ( x ) = x 3 โ 3 x + 1 on [ โ 2 , 2 ] [-2, 2] [ โ 2 , 2 ] .
Step 1: Find critical points
f โฒ ( x ) = 3 x 2 โ 3 = 3 ( x 2 โ 1 ) = 3 ( x โ 1 ) ( x + 1 ) f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1) f โฒ ( x ) = 3 x 2 โ 3 = 3 ( x 2 โ 1 ) = 3 ( x โ 1 ) ( x + 1 )
f โฒ ( x ) = 0 f'(x) = 0 f โฒ ( x ) = 0 when x = 1 x = 1 x = 1 or x = โ 1 x = -1 x = โ 1
Both are in ( โ 2 , 2 ) (-2, 2) ( โ 2 , 2 ) โ
Step 2: Evaluate at critical points
f ( โ 1 ) = ( โ 1 ) 3 โ 3 ( โ 1 ) + 1 = โ 1 + 3 + 1 = 3 f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3 f ( โ 1 ) = ( โ 1 ) 3 โ 3 ( โ 1 ) + 1 = โ 1 + 3 + 1 = 3
f ( 1 ) = 1 3 โ 3 ( 1 ) + 1 = 1 โ 3 + 1 = โ 1 f(1) = 1^3 - 3(1) + 1 = 1 - 3 + 1 = -1 f ( 1 ) = 1 3 โ 3 ( 1 ) + 1 = 1 โ 3 + 1 = โ 1
Step 3: Evaluate at endpoints
f ( โ 2 ) = ( โ 2 ) 3 โ 3 ( โ 2 ) + 1 = โ 8 + 6 + 1 = โ 1 f(-2) = (-2)^3 - 3(-2) + 1 = -8 + 6 + 1 = -1 f ( โ 2 ) = ( โ 2 ) 3 โ 3 ( โ 2 ) + 1 = โ 8 + 6 + 1 = โ 1
f ( 2 ) = 2 3 โ 3 ( 2 ) + 1 = 8 โ 6 + 1 = 3 f(2) = 2^3 - 3(2) + 1 = 8 - 6 + 1 = 3 f ( 2 ) = 2 3 โ 3 ( 2 ) + 1 = 8 โ 6 + 1 = 3
Values: f ( โ 2 ) = โ 1 f(-2) = -1 f ( โ 2 ) = โ 1 , f ( โ 1 ) = 3 f(-1) = 3 f ( โ 1 ) = 3 , f ( 1 ) = โ 1 f(1) = -1 f ( 1 ) = โ 1 , f ( 2 ) = 3 f(2) = 3 f ( 2 ) = 3
Absolute maximum: 3 3 3 at x = โ 1 x = -1 x = โ 1 and x = 2 x = 2 x = 2
Absolute minimum: โ 1 -1 โ 1 at x = 1 x = 1 x = 1 and x = โ 2 x = -2 x = โ 2
Common Types of Critical Points Type 1: Horizontal Tangent (f โฒ ( c ) = 0 f'(c) = 0 f โฒ ( c ) = 0 ) The most common type - the derivative equals zero.
Tops of hills (local max)
Bottoms of valleys (local min)
Inflection points with horizontal tangent
Type 2: Sharp Corner The function is continuous but not differentiable.
Example: f ( x ) = โฃ x โฃ f(x) = |x| f ( x ) = โฃ x โฃ at x = 0 x = 0 x = 0
f f f is continuous at x = 0 x = 0 x = 0
f โฒ ( 0 ) f'(0) f โฒ ( 0 ) does not exist (sharp turn)
x = 0 x = 0 x = 0 is a critical point (absolute minimum!)
Type 3: Vertical Tangent Example: f ( x ) = x 1 / 3 f(x) = x^{1/3} f ( x ) = x 1/3 at x = 0 x = 0 x = 0
f โฒ ( x ) = 1 3 x โ 2 / 3 = 1 3 x 2 / 3 f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}} f โฒ ( x ) = 3 1 โ x โ 2/3 = 3 x 2/3 1 โ
f โฒ ( 0 ) f'(0) f โฒ ( 0 ) is undefined (vertical tangent)
x = 0 x = 0 x = 0 is a critical point (inflection point)
Type 4: Discontinuity If f f f is discontinuous at x = c x = c x = c , it's NOT a critical point (must be in domain).
โ ๏ธ Common Mistakes
Mistake 1: Assuming All Critical Points Are Extrema โ Critical points might be inflection points or saddle points
โ
Always verify using a test (First Derivative Test, Second Derivative Test)
Mistake 2: Forgetting Endpoints When finding absolute extrema on [ a , b ] [a, b] [ a , b ] , always check the endpoints!
Mistake 3: Domain Errors x = c x = c x = c is only a critical point if it's in the domain of f f f .
Mistake 4: Undefined vs. Zero f โฒ ( c ) = 0 f'(c) = 0 f โฒ ( c ) = 0 and f โฒ ( c ) f'(c) f โฒ ( c ) undefined are both critical points, but they're different!
Quick Reference
Critical Point Checklist Absolute Extrema Checklist (on [ a , b ] [a, b] [ a , b ] )
๐ Practice Tips
Always find the derivative first before looking for critical points
Set derivative equal to zero and solve carefully
Check for undefined points in the derivative
Verify domain - make sure critical points are where f f f exists
For closed intervals , don't forget the endpoints!
To classify critical points, use First or Second Derivative Test (next lessons!)
4
x 3
+
3
๐ก Show Solution Step 1: Find the derivative
f ( x ) = x 4 โ 4 x 3 + 3 f(x) = x^4 - 4x^3 + 3 f ( x ) = x 4 โ 4 x 3 + 3
f โฒ ( x ) = 4 x 3 โ 12 x 2 f'(x) = 4x^3 - 12x^2 f โฒ ( x ) = 4 x 3 โ 12 x 2
Step 2: Factor the derivative
f โฒ ( x ) = 4 x 3 โ 12 x 2 = 4 x 2 ( x โ 3 ) f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3) f โฒ ( x ) = 4 x 3 โ 12 x
Step 3: Set equal to zero
4 x 2 ( x โ 3 ) = 0 4x^2(x - 3) = 0 4 x 2 ( x โ 3 ) = 0
This gives us:
4 x 2 = 0 4x^2 = 0 4 x 2 = 0 โ x = 0 x = 0 x = 0
x โ 3 = 0 x - 3 = 0 x โ 3 = โ
Step 4: Check if undefined
f โฒ ( x ) = 4 x 2 ( x โ 3 ) f'(x) = 4x^2(x-3) f โฒ ( x ) = 4 x 2 ( x โ 3 ) is a polynomial, so it's defined everywhere โ
Step 5: Verify in domain
Both x = 0 x = 0 x = 0 and x = 3 x = 3 x = 3 are in the domain of f f f โ
Answer : Critical points at x = 0 x = 0 x = 0 and x = 3 x = 3 x = 3
2 Problem 2medium โ Question:Find all critical points of f ( x ) = x 3 โ 6 x 2 + 9 x + 1 f(x) = x^3 - 6x^2 + 9x + 1 f ( x ) = x 3 โ 6 x 2 + 9 x + 1 .
๐ก Show Solution Solution:
Critical points occur where f โฒ ( x ) = 0 f'(x) = 0 f โฒ ( x ) = 0 or f โฒ ( x ) f'(x) f is undefined.
3 Problem 3medium โ Question:Find the absolute maximum and minimum values of g ( x ) = 2 x 3 + 3 x 2 โ 12 x + 5 g(x) = 2x^3 + 3x^2 - 12x + 5 g ( x ) = 2 x 3 + 3 x 2 โ 12 x + 5 on the interval [ โ 3 , 2 ] [-3, 2] [ โ 3 , 2 ] .
๐ก Show Solution Step 1: Find the derivative
g โฒ ( x ) = 6 x 2 + 6 x โ 12 g'(x) = 6x^2 + 6x - 12 g โฒ ( x ) = 6 x 2 + 6 x
4 Problem 4hard โ Question:Find the absolute maximum and minimum values of f ( x ) = x 3 โ 3 x 2 โ 9 x + 5 f(x) = x^3 - 3x^2 - 9x + 5 f ( x ) = x 3 โ 3 x 2 โ 9 x + 5 on the interval [ โ 2 , 4 ] [-2, 4] [ โ 2 , 4 ] .
๐ก Show Solution Solution:
Step 1: Find critical points in [ โ 2 , 4 ] [-2, 4] [ โ 2 , 4 ] .
f โฒ ( x ) = 3 x 2 โ 6 x โ 9 = 3 ( x 2 โ 2 x โ 3 ) = 3 ( x โ 3 ) ( x + 1 ) f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1) f
5 Problem 5hard โ Question:Find all critical points of h ( x ) = x 2 / 3 ( x โ 5 ) h(x) = x^{2/3}(x - 5) h ( x ) = x 2/3 ( x โ 5 ) .
๐ก Show Solution Step 1: Expand the function
h ( x ) = x 2 / 3 ( x โ 5 ) = x 5 / 3 โ 5 x 2 / 3 h(x) = x^{2/3}(x - 5) = x^{5/3} - 5x^{2/3} h ( x ) = x 2/3 ( x โ 5 ) =
โพ
Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
2
=
4 x 2 ( x โ
3 )
0
โฒ
(
x
)
f โฒ ( x ) = 3 x 2 โ 12 x + 9 f'(x) = 3x^2 - 12x + 9 f โฒ ( x ) = 3 x 2 โ 12 x + 9
3 x 2 โ 12 x + 9 = 0 3x^2 - 12x + 9 = 0 3 x 2 โ 12 x + 9 = 0
3 ( x 2 โ 4 x + 3 ) = 0 3(x^2 - 4x + 3) = 0 3 ( x 2 โ 4 x + 3 ) = 0
3 ( x โ 1 ) ( x โ 3 ) = 0 3(x - 1)(x - 3) = 0 3 ( x โ 1 ) ( x โ 3 ) = 0
x = 1 x = 1 x = 1 or x = 3 x = 3 x = 3
Find corresponding y y y -values:
f ( 1 ) = 1 โ 6 + 9 + 1 = 5 f(1) = 1 - 6 + 9 + 1 = 5 f ( 1 ) = 1 โ 6 + 9 + 1 = 5
f ( 3 ) = 27 โ 54 + 27 + 1 = 1 f(3) = 27 - 54 + 27 + 1 = 1 f ( 3 ) = 27 โ 54 + 27 + 1 = 1
Critical points: ( 1 , 5 ) (1, 5) ( 1 , 5 ) and ( 3 , 1 ) (3, 1) ( 3 , 1 )
โ
12
Step 2: Find critical points
6 x 2 + 6 x โ 12 = 0 6x^2 + 6x - 12 = 0 6 x 2 + 6 x โ 12 = 0
6 ( x 2 + x โ 2 ) = 0 6(x^2 + x - 2) = 0 6 ( x 2 + x โ 2 ) = 0
6 ( x + 2 ) ( x โ 1 ) = 0 6(x + 2)(x - 1) = 0 6 ( x + 2 ) ( x โ 1 ) = 0
So x = โ 2 x = -2 x = โ 2 or x = 1 x = 1 x = 1
Both are in [ โ 3 , 2 ] [-3, 2] [ โ 3 , 2 ] โ
Step 3: Evaluate at critical points
g ( โ 2 ) = 2 ( โ 2 ) 3 + 3 ( โ 2 ) 2 โ 12 ( โ 2 ) + 5 g(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) + 5 g ( โ 2 ) = 2 ( โ 2 ) 3 + 3 ( โ 2 ) 2 โ 12 ( โ 2 ) + 5
= 2 ( โ 8 ) + 3 ( 4 ) + 24 + 5 = 2(-8) + 3(4) + 24 + 5 = 2 ( โ 8 ) + 3 ( 4 ) + 24 + 5
= โ 16 + 12 + 24 + 5 = 25 = -16 + 12 + 24 + 5 = 25 = โ 16 + 12 + 24 + 5 = 25
g ( 1 ) = 2 ( 1 ) 3 + 3 ( 1 ) 2 โ 12 ( 1 ) + 5 g(1) = 2(1)^3 + 3(1)^2 - 12(1) + 5 g ( 1 ) = 2 ( 1 ) 3 + 3 ( 1 ) 2 โ 12 ( 1 ) + 5
= 2 + 3 โ 12 + 5 = โ 2 = 2 + 3 - 12 + 5 = -2 = 2 + 3 โ 12 + 5 = โ 2
Step 4: Evaluate at endpoints
g ( โ 3 ) = 2 ( โ 3 ) 3 + 3 ( โ 3 ) 2 โ 12 ( โ 3 ) + 5 g(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) + 5 g ( โ 3 ) = 2 ( โ 3 ) 3 + 3 ( โ 3 ) 2 โ 12 ( โ 3 ) + 5
= 2 ( โ 27 ) + 3 ( 9 ) + 36 + 5 = 2(-27) + 3(9) + 36 + 5 = 2 ( โ 27 ) + 3 ( 9 ) + 36 + 5
= โ 54 + 27 + 36 + 5 = 14 = -54 + 27 + 36 + 5 = 14 = โ 54 + 27 + 36 + 5 = 14
g ( 2 ) = 2 ( 2 ) 3 + 3 ( 2 ) 2 โ 12 ( 2 ) + 5 g(2) = 2(2)^3 + 3(2)^2 - 12(2) + 5 g ( 2 ) = 2 ( 2 ) 3 + 3 ( 2 ) 2 โ 12 ( 2 ) + 5
= 2 ( 8 ) + 3 ( 4 ) โ 24 + 5 = 2(8) + 3(4) - 24 + 5 = 2 ( 8 ) + 3 ( 4 ) โ 24 + 5
= 16 + 12 โ 24 + 5 = 9 = 16 + 12 - 24 + 5 = 9 = 16 + 12 โ 24 + 5 = 9
Step 5: Compare all values
g ( โ 3 ) = 14 g(-3) = 14 g ( โ 3 ) = 14 , g ( โ 2 ) = 25 g(-2) = 25 g ( โ 2 ) = 25 , g ( 1 ) = โ 2 g(1) = -2 g ( 1 ) = โ 2 , g ( 2 ) = 9 g(2) = 9 g ( 2 ) = 9
Absolute maximum : 25 25 25 at x = โ 2 x = -2 x = โ 2
Absolute minimum : โ 2 -2 โ 2 at x = 1 x = 1 x = 1
โฒ
(
x
)
=
3 x 2 โ
6 x โ
9 =
3 ( x 2 โ
2 x โ
3 ) =
3 ( x โ
3 ) ( x +
1 )
Critical points: x = 3 x = 3 x = 3 and x = โ 1 x = -1 x = โ 1 (both in the interval)
Step 2: Evaluate f f f at critical points and endpoints.
f ( โ 2 ) = ( โ 2 ) 3 โ 3 ( โ 2 ) 2 โ 9 ( โ 2 ) + 5 = โ 8 โ 12 + 18 + 5 = 3 f(-2) = (-2)^3 - 3(-2)^2 - 9(-2) + 5 = -8 - 12 + 18 + 5 = 3 f ( โ 2 ) = ( โ 2 ) 3 โ 3 ( โ 2 ) 2 โ 9 ( โ 2 ) + 5 = โ 8 โ 12 + 18 + 5 = 3
f ( โ 1 ) = ( โ 1 ) 3 โ 3 ( โ 1 ) 2 โ 9 ( โ 1 ) + 5 = โ 1 โ 3 + 9 + 5 = 10 f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10 f ( โ 1 ) = ( โ 1 ) 3 โ 3 ( โ 1 ) 2 โ 9 ( โ 1 ) + 5 = โ 1 โ 3 + 9 + 5 = 10
f ( 3 ) = ( 3 ) 3 โ 3 ( 3 ) 2 โ 9 ( 3 ) + 5 = 27 โ 27 โ 27 + 5 = โ 22 f(3) = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22 f ( 3 ) = ( 3 ) 3 โ 3 ( 3 ) 2 โ 9 ( 3 ) + 5 = 27 โ 27 โ 27 + 5 = โ 22
f ( 4 ) = ( 4 ) 3 โ 3 ( 4 ) 2 โ 9 ( 4 ) + 5 = 64 โ 48 โ 36 + 5 = โ 15 f(4) = (4)^3 - 3(4)^2 - 9(4) + 5 = 64 - 48 - 36 + 5 = -15 f ( 4 ) = ( 4 ) 3 โ 3 ( 4 ) 2 โ 9 ( 4 ) + 5 = 64 โ 48 โ 36 + 5 = โ 15
Absolute maximum: f ( โ 1 ) = 10 f(-1) = 10 f ( โ 1 ) = 10
Absolute minimum: f ( 3 ) = โ 22 f(3) = -22 f ( 3 ) = โ 22
x
5/3
โ
5 x 2/3
Step 2: Find the derivative
h โฒ ( x ) = 5 3 x 2 / 3 โ 5 โ
2 3 x โ 1 / 3 h'(x) = \frac{5}{3}x^{2/3} - 5 \cdot \frac{2}{3}x^{-1/3} h โฒ ( x ) = 3 5 โ x 2/3 โ 5 โ
3 2 โ x โ 1/3
h โฒ ( x ) = 5 3 x 2 / 3 โ 10 3 x โ 1 / 3 h'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} h โฒ ( x ) = 3 5 โ x 2/3 โ 3 10 โ x โ 1/3
Step 3: Factor using common denominator
h โฒ ( x ) = 5 x 2 / 3 3 โ 10 3 x 1 / 3 h'(x) = \frac{5x^{2/3}}{3} - \frac{10}{3x^{1/3}} h โฒ ( x ) = 3 5 x 2/3 โ โ 3 x 1/3 10 โ
h โฒ ( x ) = 5 x 2 / 3 โ
x 1 / 3 โ 10 3 x 1 / 3 h'(x) = \frac{5x^{2/3} \cdot x^{1/3} - 10}{3x^{1/3}} h โฒ ( x ) = 3 x 1/3 5 x 2/3 โ
x 1/3 โ 10 โ
h โฒ ( x ) = 5 x โ 10 3 x 1 / 3 h'(x) = \frac{5x - 10}{3x^{1/3}} h โฒ ( x ) = 3 x 1/3 5 x โ 10 โ
h โฒ ( x ) = 5 ( x โ 2 ) 3 x 1 / 3 h'(x) = \frac{5(x - 2)}{3x^{1/3}} h โฒ ( x ) = 3 x 1/3 5 ( x โ 2 ) โ
Step 4: Find where h โฒ ( x ) = 0 h'(x) = 0 h โฒ ( x ) = 0
Numerator = 0: 5 ( x โ 2 ) = 0 5(x - 2) = 0 5 ( x โ 2 ) = 0
Step 5: Find where h โฒ ( x ) h'(x) h โฒ ( x ) is undefined
Denominator = 0: 3 x 1 / 3 = 0 3x^{1/3} = 0 3 x 1/3 = 0
h ( 0 ) = 0 2 / 3 ( 0 โ 5 ) = 0 h(0) = 0^{2/3}(0 - 5) = 0 h ( 0 ) = 0 2/3 ( 0 โ 5 ) = 0 (in domain) โ
Answer : Critical points at x = 0 x = 0 x = 0 and x = 2 x = 2 x = 2
Note : At x = 0 x = 0 x = 0 , there's a vertical tangent (cusp).