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.
Curve Sketching | Study Mondo
Topics / Applications of Derivatives / Curve Sketching Curve Sketching Using derivatives to sketch accurate graphs 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 Curve Sketching
What is Curve Sketching?
Curve sketching is the art of drawing an accurate graph of a function using calculus. Instead of plotting points, we use derivatives to understand the function's behavior!
๐ก Key Idea : Use f โฒ ( x ) f'(x) f โฒ ( x ) and f โฒ โฒ ( x ) f''(x) f to determine where the function is increasing/decreasing and where it's curving up/down.
๐ Practice Problems
1 Problem 1medium โ Question:Sketch the graph of f ( x ) = x 4 โ 4 x 3 f(x) = x^4 - 4x^3 f ( x ) = x 4 โ using calculus.
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: Curve SketchingAvoid 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: Curve SketchingSee 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 Curve Sketching
โพ ๐ Related Topics in Applications of Derivativesโ Frequently Asked QuestionsWhat is Curve Sketching?โพ Using derivatives to sketch accurate graphs of functions
How can I study Curve Sketching 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 Curve Sketching study guide free?โพ Yes โ all study notes, flashcards, and practice problems for Curve Sketching on Study Mondo are 100% free. No account is needed to access the content.
What course covers Curve Sketching?โพ Curve Sketching 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 Curve Sketching?
๐ก Study Tipsโ Work through examples step-by-step โ Practice with flashcards daily โ Review common mistakes โฒโฒ
(
x
)
The Complete Curve Sketching Checklist
Step-by-Step Guide
y y y -intercept: Set x = 0 x = 0 x = 0
x x x -intercepts: Solve f ( x ) = 0 f(x) = 0 f ( x ) = 0
Step 3 : Check for symmetry
Even: f ( โ x ) = f ( x ) f(-x) = f(x) f ( โ x ) = f ( x ) (symmetric about y y y -axis)
Odd: f ( โ x ) = โ f ( x ) f(-x) = -f(x) f ( โ x ) = โ f ( x ) (symmetric about origin)
Vertical: Where denominator = 0 (for rational functions)
Horizontal: Check lim โก x โ ยฑ โ f ( x ) \lim_{x \to \pm\infty} f(x) lim x โ ยฑ โ โ f ( x )
Slant: If degree of numerator is 1 more than denominator
Step 5 : Find critical points (f โฒ ( x ) = 0 f'(x) = 0 f โฒ ( x ) = 0 or undefined)
Step 6 : Determine intervals of increase/decrease
f โฒ ( x ) > 0 f'(x) > 0 f โฒ ( x ) > 0 : increasing
f โฒ ( x ) < 0 f'(x) < 0 f โฒ ( x ) < 0 : decreasing
Step 7 : Find local extrema (use First or Second Derivative Test)
Step 8 : Find inflection points (f โฒ โฒ ( x ) = 0 f''(x) = 0 f โฒโฒ ( x ) = 0 and concavity changes)
Step 9 : Determine concavity
f โฒ โฒ ( x ) > 0 f''(x) > 0 f โฒโฒ ( x ) > 0 : concave up โช
f โฒ โฒ ( x ) < 0 f''(x) < 0 f โฒโฒ ( x ) < 0 : concave down โฉ
Step 10 : Sketch the graph using all information!
What Each Derivative Tells You First Derivative: f โฒ ( x ) f'(x) f โฒ ( x ) Sign of f โฒ ( x ) f'(x) f โฒ ( x ) :
f โฒ ( x ) > 0 f'(x) > 0 f โฒ ( x ) > 0 โ function increasing โ
f โฒ ( x ) < 0 f'(x) < 0 f โฒ ( x ) < 0 โ function decreasing โ
f โฒ ( x ) = 0 f'(x) = 0 f โฒ ( x ) = 0 โ horizontal tangent (potential max/min)
Critical points : Where f โฒ ( x ) = 0 f'(x) = 0 f โฒ ( x ) = 0 or f โฒ ( x ) f'(x) f โฒ ( x ) undefined
Local extrema : Where f โฒ f' f โฒ changes sign
Second Derivative: f โฒ โฒ ( x ) f''(x) f โฒโฒ ( x ) Sign of f โฒ โฒ ( x ) f''(x) f โฒโฒ ( x ) :
f โฒ โฒ ( x ) > 0 f''(x) > 0 f โฒโฒ ( x ) > 0 โ concave up โช (curving upward)
f โฒ โฒ ( x ) < 0 f''(x) < 0 f โฒโฒ ( x ) < 0 โ concave down โฉ (curving downward)
f โฒ โฒ ( x ) = 0 f''(x) = 0 f โฒโฒ ( x ) = 0 โ potential inflection point
Inflection points : Where concavity changes (and f โฒ โฒ f'' f โฒโฒ changes sign)
Example: Complete Curve Sketch Sketch 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
Polynomial โ domain is all real numbers: ( โ โ , โ ) (-\infty, \infty) ( โ โ , โ )
y y y -intercept: f ( 0 ) = 1 f(0) = 1 f ( 0 ) = 1 โ point ( 0 , 1 ) (0, 1) ( 0 , 1 )
x x x -intercepts: x 3 โ 6 x 2 + 9 x + 1 = 0 x^3 - 6x^2 + 9x + 1 = 0 x 3 โ 6 x 2 + 9 x + 1 = 0 (hard to solve, skip for now)
f ( โ x ) = โ x 3 โ 6 x 2 โ 9 x + 1 โ f ( x ) f(-x) = -x^3 - 6x^2 - 9x + 1 \neq f(x) f ( โ x ) = โ x 3 โ 6 x 2 โ 9 x + 1 ๎ = f ( x ) and โ โ f ( x ) \neq -f(x) ๎ = โ f ( x )
Polynomial โ no asymptotes
Step 5: First derivative and critical points
f โฒ ( x ) = 3 x 2 โ 12 x + 9 = 3 ( x 2 โ 4 x + 3 ) = 3 ( x โ 1 ) ( x โ 3 ) f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3) f โฒ ( x ) = 3 x 2 โ 12 x + 9 = 3 ( x 2 โ 4 x + 3 ) = 3 ( x โ 1 ) ( x โ 3 )
Critical points: x = 1 x = 1 x = 1 and x = 3 x = 3 x = 3
Step 6: Sign chart for f โฒ ( x ) f'(x) f โฒ ( x )
1 3
++++ | ---- | ++++
โ โผ โ โฒ โ
Increasing on ( โ โ , 1 ) (-\infty, 1) ( โ โ , 1 ) and ( 3 , โ ) (3, \infty) ( 3 , โ )
Decreasing on ( 1 , 3 ) (1, 3) ( 1 , 3 )
At x = 1 x = 1 x = 1 : f โฒ f' f โฒ changes + to โ โ local max
f ( 1 ) = 1 โ 6 + 9 + 1 = 5 f(1) = 1 - 6 + 9 + 1 = 5 f ( 1 ) = 1 โ 6 + 9 + 1 = 5
At x = 3 x = 3 x = 3 : f โฒ f' f โฒ changes โ to + โ local min
f ( 3 ) = 27 โ 54 + 27 + 1 = 1 f(3) = 27 - 54 + 27 + 1 = 1 f ( 3 ) = 27 โ 54 + 27 + 1 = 1
Step 8: Second derivative and inflection points
f โฒ โฒ ( x ) = 6 x โ 12 = 6 ( x โ 2 ) f''(x) = 6x - 12 = 6(x - 2) f โฒโฒ ( x ) = 6 x โ 12 = 6 ( x โ 2 )
f โฒ โฒ ( x ) = 0 f''(x) = 0 f โฒโฒ ( x ) = 0 when x = 2 x = 2 x = 2
Concave down on ( โ โ , 2 ) (-\infty, 2) ( โ โ , 2 )
Concave up on ( 2 , โ ) (2, \infty) ( 2 , โ )
Inflection point at x = 2 x = 2 x = 2 : f ( 2 ) = 8 โ 24 + 18 + 1 = 3 f(2) = 8 - 24 + 18 + 1 = 3 f ( 2 ) = 8 โ 24 + 18 + 1 = 3
( 0 , 1 ) (0, 1) ( 0 , 1 ) - y y y -intercept
( 1 , 5 ) (1, 5) ( 1 , 5 ) - local max
( 2 , 3 ) (2, 3) ( 2 , 3 ) - inflection point
( 3 , 1 ) (3, 1) ( 3 , 1 ) - local min
Increases to ( 1 , 5 ) (1, 5) ( 1 , 5 ) , then decreases to ( 3 , 1 ) (3, 1) ( 3 , 1 ) , then increases again
Concave down until x = 2 x = 2 x = 2 , then concave up
Passes through ( 0 , 1 ) (0, 1) ( 0 , 1 )
Special Features to Look For
Cusps and Corners Where f f f is continuous but f โฒ f' f โฒ doesn't exist
Example : f ( x ) = โฃ x โฃ f(x) = |x| f ( x ) = โฃ x โฃ has a corner at x = 0 x = 0 x = 0
Vertical Tangents Where f โฒ ( x ) = ยฑ โ f'(x) = \pm\infty f โฒ ( x ) = ยฑ โ
Example : f ( x ) = x 1 / 3 f(x) = x^{1/3} f ( x ) = x 1/3 has vertical tangent at x = 0 x = 0 x = 0
Discontinuities
Jump discontinuity
Removable discontinuity
Infinite discontinuity (vertical asymptote)
Analyzing Rational Functions For f ( x ) = p ( x ) q ( x ) f(x) = \frac{p(x)}{q(x)} f ( x ) = q ( x ) p ( x ) โ :
Vertical Asymptotes Occur where q ( x ) = 0 q(x) = 0 q ( x ) = 0 (denominator zero, numerator non-zero)
Example : f ( x ) = 1 x โ 2 f(x) = \frac{1}{x-2} f ( x ) = x โ 2 1 โ has vertical asymptote at x = 2 x = 2 x = 2
Horizontal Asymptotes Compare degrees of p ( x ) p(x) p ( x ) and q ( x ) q(x) q ( x ) :
If deg(p p p ) < deg(q q q ): y = 0 y = 0 y = 0 is horizontal asymptote
If deg(p p p ) = deg(q q q ): y = leadingย coefย ofย p leadingย coefย ofย q y = \frac{\text{leading coef of } p}{\text{leading coef of } q} y = leadingย coefย ofย q leadingย coefย ofย p โ
If deg(p p p ) > deg(q q q ): No horizontal asymptote (may have slant asymptote)
Example : f ( x ) = 2 x 2 + 1 x 2 โ 4 f(x) = \frac{2x^2 + 1}{x^2 - 4} f ( x ) = x 2 โ 4 2 x 2 + 1 โ has horizontal asymptote y = 2 y = 2 y = 2
Slant (Oblique) Asymptotes When deg(p p p ) = deg(q q q ) + 1, use polynomial division
Example : f ( x ) = x 2 + 1 x f(x) = \frac{x^2 + 1}{x} f ( x ) = x x 2 + 1 โ has slant asymptote y = x y = x y = x
Quick Summary Table Feature How to Find What It Tells You Domain Where f f f is defined Valid x x x values Intercepts f ( 0 ) f(0) f ( 0 ) and f ( x ) = 0 f(x) = 0 f ( x ) = 0 Where graph crosses axes Critical pts f โฒ ( x ) = 0 f'(x) = 0 f โฒ ( x ) = 0 or undefPotential max/min Inc/Dec Sign of f โฒ ( x ) f'(x) f โฒ ( x ) Direction of graph Local extrema Sign change in f โฒ f' f โฒ Peaks and valleys Inflection pts f โฒ โฒ ( x ) = 0 f''(x) = 0 f โฒโฒ ( x ) = 0 + sign changeWhere concavity changes Concavity Sign of f โฒ โฒ ( x ) f''(x) f โฒโฒ ( x ) Curvature direction Asymptotes Limits, denominators Boundary behavior
โ ๏ธ Common Mistakes
Mistake 1: Not Checking Sign Changes f โฒ โฒ ( c ) = 0 f''(c) = 0 f โฒโฒ ( c ) = 0 doesn't guarantee an inflection point - concavity must change !
Mistake 2: Forgetting Domain Always consider where the function is actually defined.
Mistake 3: Plotting Only Critical Points Include inflection points, intercepts, and other key features!
Mistake 4: Wrong Asymptote Analysis Vertical asymptotes: denominator = 0
Horizontal asymptotes: check limits at infinity
Mistake 5: Ignoring Behavior at Infinity Always check what happens as x โ ยฑ โ x \to \pm\infty x โ ยฑ โ
The Big Picture
Putting It All Together
f ( x ) f(x) f ( x ) tells you the height of the graph
f โฒ ( x ) f'(x) f โฒ ( x ) tells you the slope (increasing/decreasing)
f โฒ โฒ ( x ) f''(x) f โฒโฒ ( x ) tells you the curvature (concave up/down)
All three work together to give you a complete picture!
Shortcut for Simple Polynomials For polynomials, you can often skip some steps:
No asymptotes
Domain is always R \mathbb{R} R
Continuous everywhere
Focus on critical points, extrema, and inflection points
๐ Practice Strategy
Follow the checklist systematically
Make a sign chart for both f โฒ f' f โฒ and f โฒ โฒ f'' f โฒโฒ
Calculate key points (don't just mark x x x -values, find y y y too!)
Sketch lightly first , then refine
Check your work : Does the sketch match your analysis?
Label everything : maxima, minima, inflection points, asymptotes
4
x 3
๐ก Show Solution Step 1: Domain
Polynomial โ ( โ โ , โ ) (-\infty, \infty) ( โ โ , โ )
Step 2: Intercepts
y y y -intercept: f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 โ ( 0 , 0 ) (0, 0) ( 0 , 0 )
x x x -intercepts: x 4 โ 4 x 3 = 0 x^4 - 4x^3 = 0 x 4 โ 4 x 3 = 0
x 3 ( x โ 4 ) = 0 x^3(x - 4) = 0
or โ and
Step 3: Symmetry
f ( โ x ) = x 4 + 4 x 3 โ f ( x ) f(-x) = x^4 + 4x^3 \neq f(x) f ( โ x ) = x 4 + 4 x 3 ๎ = โ No symmetry
Step 4: First 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
Critical points: x = 0 , 3 x = 0, 3 x = 0 , 3
Step 5: Sign chart for f โฒ f' f โฒ
Test points:
x = โ 1 x = -1 x = โ 1 : f โฒ ( โ 1 ) = 4 ( 1 ) ( โ 4 ) = โ 16 < 0 f'(-1) = 4(1)(-4) = -16 < 0 f โฒ ( โ 1 ) = 4 ( 1 ) ( โ 4 )
0 3
---- | ---- | ++++
โ โ โฒ โ
Decreasing on ( โ โ , 3 ) (-\infty, 3) ( โ โ , 3 ) , increasing on ( 3 , โ ) (3, \infty) ( 3 , โ )
Local minimum at x = 3 x = 3 x = 3 : f ( 3 ) = 81 โ 108 = โ 27 f(3) = 81 - 108 = -27 f ( 3 ) = 81 โ 108 = โ 27
Note: x = 0 x = 0 x = 0 is NOT an extremum (no sign change)
Step 6: Second derivative
f โฒ โฒ ( x ) = 12 x 2 โ 24 x = 12 x ( x โ 2 ) f''(x) = 12x^2 - 24x = 12x(x - 2) f โฒโฒ ( x ) = 12 x 2 โ 24 x = 12 x (
f โฒ โฒ ( x ) = 0 f''(x) = 0 f โฒโฒ ( x ) = 0 when x = 0 x = 0 x = 0 or x = 2 x = 2 x =
Step 7: Concavity
Test points:
x = โ 1 x = -1 x = โ 1 : f โฒ โฒ ( โ 1 ) = 12 ( 1 ) ( โ 3 ) = โ 36... w a i t , l e t m e r e c a l c u l a t e f''(-1) = 12(1)(-3) = -36... wait, let me recalculate f โฒโฒ ( โ 1 )
0 2
++++ | ---- | ++++
โช โฉ โช
Inflection points at x = 0 x = 0 x = 0 and x = 2 x = 2 x = 2
f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 , f ( 2 ) = 16 โ 32 = โ 16 f(2) = 16 - 32 = -16 f ( 2 ) = 16 โ 32 = โ 16
Summary for sketch :
Key points:
( 0 , 0 ) (0, 0) ( 0 , 0 ) - intercept and inflection point
( 2 , โ 16 ) (2, -16) ( 2 , โ 16 ) - inflection point
( 3 , โ 27 ) (3, -27) ( 3 , โ 27 ) - local minimum
( 4 , 0 - -intercept
Behavior:
Decreasing from โ โ -\infty โ โ to ( 3 , โ 27 ) (3, -27) ( 3 , โ 27 )
Then increasing to + โ +\infty + โ
Concave up, then down, then up again
Passes through origin
2 Problem 2hard โ Question:Sketch the graph of g ( x ) = x 2 x 2 โ 4 g(x) = \frac{x^2}{x^2 - 4} g ( x ) = x 2 โ 4 x 2 โ using calculus. Include all asymptotes.
๐ก Show Solution Step 1: Domain
Denominator: x 2 โ 4 = ( x โ 2 ) ( x + 2 ) = 0 x^2 - 4 = (x-2)(x+2) = 0 x 2 โ 4 = ( x โ 2 ) ( x + when
3 Problem 3expert โ Question:Analyze and sketch h ( x ) = x e โ x h(x) = xe^{-x} h ( x ) = x e โ x completely.
๐ก Show Solution Step 1: Domain
All real numbers: ( โ โ , โ ) (-\infty, \infty) ( โ โ , โ )
Step 2: Intercepts
y y y -intercept: h ( 0 ) = 0 โ
e 0 = 0 h(0) = 0 \cdot e^0 = 0 h ( 0 ) = 0 โ
e 0 = 0 โ ( 0 , 0 ) (0, 0)
x x x -intercepts: x e โ x = 0 xe^{-x} = 0 x e โ x = 0 โ x = 0 x = 0 x = 0 โ ( 0 , 0 ) (0, 0)
Step 3: Limits (behavior at infinity)
As x โ โ โ x \to -\infty x โ โ โ : e โ x โ โ e^{-x} \to \infty e โ x โ โ , so h ( x ) โ โ โ h(x) \to -\infty h ( x )
As x โ + โ x \to +\infty x โ + โ : x e x โ 0 \frac{x}{e^x} \to 0 e x x โ โ (exponential dominates)
Horizontal asymptote: y = 0 y = 0 y = 0 (as x โ โ x \to \infty x โ โ )
Step 4: First derivative (Product Rule)
h โฒ ( x ) = ( 1 ) ( e โ x ) + ( x ) ( โ e โ x ) h'(x) = (1)(e^{-x}) + (x)(-e^{-x}) h โฒ ( x ) = ( 1 ) ( e โ x ) + (
= e โ x โ x e โ x = e^{-x} - xe^{-x} = e โ x โ x e โ x
= e โ x ( 1 โ x ) = e^{-x}(1 - x) = e โ x ( 1 โ x )
Critical point: 1 โ x = 0 1 - x = 0 1 โ x = 0 โ x = 1 x = 1 x = 1
Step 5: Sign of h โฒ h' h โฒ
e โ x > 0 e^{-x} > 0 e โ x > 0 always, so sign depends on ( 1 โ x ) (1-x) ( 1 โ x ) :
x < 1 x < 1 x < 1 : ( 1 โ x ) > 0 (1-x) > 0 ( 1 โ x ) > 0 โ increasing
x > 1 x > 1 x > 1 : โ decreasing
Local maximum at x = 1 x = 1 x = 1 : h ( 1 ) = 1 โ
e โ 1 = 1 e โ 0.368 h(1) = 1 \cdot e^{-1} = \frac{1}{e} \approx 0.368 h ( 1 ) = 1 โ
e โ 1 =
Step 6: Second derivative
h โฒ โฒ ( x ) = ( โ e โ x ) ( 1 โ x ) + e โ x ( โ 1 ) h''(x) = (-e^{-x})(1-x) + e^{-x}(-1) h โฒโฒ ( x ) = ( โ e โ x ) ( 1 โ
= โ e โ x ( 1 โ x ) โ e โ x = -e^{-x}(1-x) - e^{-x} = โ e โ x ( 1 โ x ) โ e โ x
= โ e โ x ( 1 โ x + 1 ) = -e^{-x}(1-x+1) = โ e โ x ( 1 โ x + 1 )
= โ e โ x ( 2 โ x ) = -e^{-x}(2-x) = โ e โ x ( 2 โ x )
= e โ x ( x โ 2 ) = e^{-x}(x-2) = e โ x ( x โ 2 )
Inflection point: x โ 2 = 0 x - 2 = 0 x โ 2 = 0 โ x = 2 x = 2 x = 2
Step 7: Concavity
x < 2 x < 2 x < 2 : ( x โ 2 ) < 0 (x-2) < 0 ( x โ 2 ) < 0 โ concave down
x > 2 x > 2 x > 2 : โ concave up
Inflection point at x = 2 x = 2 x = 2 : h ( 2 ) = 2 e โ 2 = 2 e 2 โ 0.271 h(2) = 2e^{-2} = \frac{2}{e^2} \approx 0.271 h ( 2 ) = 2 e โ 2 =
Summary for sketch :
Key points:
( 0 , 0 ) (0, 0) ( 0 , 0 ) - origin
( 1 , 1 / e ) (1, 1/e) ( 1 , 1/ e ) - local maximum
( 2 , 2 / e 2 ) (2, 2/e^2) ( 2 , 2/ e - inflection point
Behavior:
Increases from โ โ -\infty โ โ to max at x = 1 x=1 x = 1
Decreases from max, approaching 0 as x โ โ x \to \infty x โ โ
Concave down until x = 2 x=2 x = , then concave up
Answer : The graph rises from negative infinity through the origin, reaches a maximum at ( 1 , 1 / e ) (1, 1/e) ( 1 , 1/ e ) , has an inflection point at ( 2 , 2 / e 2 ) (2, 2/e^2) ( 2 , 2/ e 2 ) , and approaches 0 as x โ โ x \to \infty x โ .
4 Problem 4medium โ Question:Sketch the curve y = xยณ - 3xยฒ using calculus techniques.
๐ก Show Solution Step 1: Find intercepts:
y-intercept: y(0) = 0
x-intercepts: xยณ - 3xยฒ = 0 โ xยฒ(x - 3) = 0 โ x = 0, 3
Step 2: Find f'(x) and critical points:
f'(x) = 3xยฒ - 6x = 3x(x - 2)
Critical points: x = 0, 2
Step 3: First derivative test:
f'(-1) = 3(-1)(โ3) = 9 > 0 (increasing)
f'(1) = 3(1)(โ1) = -3 < 0 (decreasing)
f'(3) = 3(3)(1) = 9 > 0 (increasing)
Local max at x = 0: f(0) = 0
Local min at x = 2: f(2) = 8 - 12 = -4
Step 4: Find f''(x) and inflection points:
f''(x) = 6x - 6 = 6(x - 1)
Inflection point: x = 1, f(1) = 1 - 3 = -2
Step 5: Concavity:
f''(0) = -6 < 0 (concave down)
f''(2) = 6 > 0 (concave up)
Step 6: End behavior:
As x โ โ: y โ โ (positive cubic)
As x โ -โ: y โ -โ
Answer: Curve crosses (0,0) and (3,0), local max at (0,0), local min at (2,-4), inflection at (1,-2)
5 Problem 5hard โ Question:Analyze and sketch f(x) = x/(xยฒ + 1).
๐ก Show Solution Step 1: Domain and intercepts:
Domain: all real numbers (denominator never zero)
x-intercept: x = 0
y-intercept: f(0) = 0
Step 2: Symmetry:
f(-x) = -x/(xยฒ + 1) = -f(x) โ odd function (symmetric about origin)
Step 3: Horizontal asymptotes:
lim(xโยฑโ) x/(xยฒ + 1) = lim(xโยฑโ) (1/x)/(1 + 1/xยฒ) = 0
Horizontal asymptote: y = 0
Step 4: Find f'(x) using quotient rule:
f'(x) = [(1)(xยฒ + 1) - x(2x)]/(xยฒ + 1)ยฒ
= (xยฒ + 1 - 2xยฒ)/(xยฒ + 1)ยฒ
= (1 - xยฒ)/(xยฒ + 1)ยฒ
Step 5: Critical points:
f'(x) = 0 when 1 - xยฒ = 0 โ x = ยฑ1
f(1) = 1/2, f(-1) = -1/2
Local max at (1, 1/2), local min at (-1, -1/2)
Step 6: Find f''(x) (or just note concavity):
By symmetry and shape, inflection points exist
Answer: Odd function, passes through origin, local max (1, 1/2), local min (-1, -1/2), horizontal asymptote y = 0
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.
x 3 ( x โ
4 ) =
0
f ( x )
2
=
4 x 2 ( x โ
3 )
=
โ 16 <
0
x = 1 x = 1 x = 1 : f โฒ ( 1 ) = 4 ( 1 ) ( โ 2 ) = โ 8 < 0 f'(1) = 4(1)(-2) = -8 < 0 f โฒ ( 1 ) = 4 ( 1 ) ( โ 2 ) = โ 8 < 0 x = 4 x = 4 x = 4 : f โฒ ( 4 ) = 4 ( 16 ) ( 1 ) = 64 > 0 f'(4) = 4(16)(1) = 64 > 0 f โฒ ( 4 ) = 4 ( 16 ) ( 1 ) = 64 > 0 x
โ
2 )
2
=
12 ( 1 ) ( โ 3 ) =
โ 36... w ai t , l e t m erec a l c u l a t e
x = โ 1 x = -1 x = โ 1 : f โฒ โฒ ( โ 1 ) = 12 ( โ 1 ) ( โ 1 โ 2 ) = 12 ( โ 1 ) ( โ 3 ) = 36 > 0 f''(-1) = 12(-1)(-1-2) = 12(-1)(-3) = 36 > 0 f โฒโฒ ( โ 1 ) = 12 ( โ 1 ) ( โ 1 โ 2 ) = 12 ( โ 1 ) ( โ 3 ) = 36 > 0 x = 1 x = 1 x = 1 : f โฒ โฒ ( 1 ) = 12 ( 1 ) ( 1 โ 2 ) = 12 ( 1 ) ( โ 1 ) = โ 12 < 0 f''(1) = 12(1)(1-2) = 12(1)(-1) = -12 < 0 f โฒโฒ ( 1 ) = 12 ( 1 ) ( 1 โ 2 ) = 12 ( 1 ) ( โ 1 ) = โ 12 < 0 x = 3 x = 3 x = 3 : f โฒ โฒ ( 3 ) = 12 ( 3 ) ( 3 โ 2 ) = 36 > 0 f''(3) = 12(3)(3-2) = 36 > 0 f โฒโฒ ( 3 ) = 12 ( 3 ) ( 3 โ 2 ) = 36 > 0 ) (4, 0) ( 4 , 0 )
2
)
=
0
Domain: ( โ โ , โ 2 ) โช ( โ 2 , 2 ) โช ( 2 , โ ) (-\infty, -2) \cup (-2, 2) \cup (2, \infty) ( โ โ , โ 2 ) โช ( โ 2 , 2 ) โช ( 2 , โ )
y y y -intercept: g ( 0 ) = 0 โ 4 = 0 g(0) = \frac{0}{-4} = 0 g ( 0 ) = โ 4 0 โ = 0 โ ( 0 , 0 ) (0, 0) ( 0 , 0 )
x x x -intercepts: x 2 = 0 x^2 = 0 x 2 = 0 โ x = 0 x = 0 x = 0 โ ( 0 , 0 ) (0, 0) ( 0 , 0 )
g ( โ x ) = ( โ x ) 2 ( โ x ) 2 โ 4 = x 2 x 2 โ 4 = g ( x ) g(-x) = \frac{(-x)^2}{(-x)^2 - 4} = \frac{x^2}{x^2-4} = g(x) g ( โ x ) = ( โ x ) 2 โ 4 ( โ x ) 2 โ = x 2 โ 4 x 2 โ = g ( x )
EVEN function - symmetric about y y y -axis!
Vertical : At x = โ 2 x = -2 x = โ 2 and x = 2 x = 2 x = 2 (where denominator = 0)
Horizontal :
lim โก x โ โ x 2 x 2 โ 4 = lim โก x โ โ 1 1 โ 4 / x 2 = 1 \lim_{x \to \infty} \frac{x^2}{x^2 - 4} = \lim_{x \to \infty} \frac{1}{1 - 4/x^2} = 1 lim x โ โ โ x 2 โ 4 x 2 โ = lim x โ โ โ 1 โ 4/ x 2 1 โ = 1
Horizontal asymptote: y = 1 y = 1 y = 1
Step 5: First derivative (Quotient Rule)
g โฒ ( x ) = 2 x ( x 2 โ 4 ) โ x 2 ( 2 x ) ( x 2 โ 4 ) 2 g'(x) = \frac{2x(x^2-4) - x^2(2x)}{(x^2-4)^2} g โฒ ( x ) = ( x 2 โ 4 ) 2 2 x ( x 2 โ 4 ) โ x 2 ( 2 x ) โ
= 2 x 3 โ 8 x โ 2 x 3 ( x 2 โ 4 ) 2 = \frac{2x^3 - 8x - 2x^3}{(x^2-4)^2} = ( x 2 โ 4 ) 2 2 x 3 โ 8 x โ 2 x 3 โ
= โ 8 x ( x 2 โ 4 ) 2 = \frac{-8x}{(x^2-4)^2} = ( x 2 โ 4 ) 2 โ 8 x โ
Critical point: โ 8 x = 0 -8x = 0 โ 8 x = 0 โ x = 0 x = 0 x = 0
Step 6: Sign of g โฒ g' g โฒ
Since ( x 2 โ 4 ) 2 > 0 (x^2-4)^2 > 0 ( x 2 โ 4 ) 2 > 0 always, sign depends on โ 8 x -8x โ 8 x :
x < 0 x < 0 x < 0 : โ 8 x > 0 -8x > 0 โ 8 x > 0 โ increasing
x > 0 x > 0 x > 0 : โ 8 x < 0 -8x < 0 โ 8 x < 0 โ decreasing
Local maximum at x = 0 x = 0 x = 0 : g ( 0 ) = 0 g(0) = 0 g ( 0 ) = 0
Step 7: Second derivative (skip for brevity)
Can verify concavity, but we have enough information.
Step 8: Behavior near asymptotes
Near x = 2 + x = 2^+ x = 2 + : numerator โ 4, denominator โ 0 + 0^+ 0 + โ g ( x ) โ + โ g(x) \to +\infty g ( x ) โ + โ
Near x = 2 โ x = 2^- x = 2 โ : numerator โ 4, denominator โ 0 โ 0^- 0 โ โ g ( x ) โ โ โ g(x) \to -\infty g ( x ) โ โ โ
By symmetry, similar behavior at x = โ 2 x = -2 x = โ 2
Symmetric about y y y -axis
Vertical asymptotes at x = ยฑ 2 x = \pm 2 x = ยฑ 2
Horizontal asymptote at y = 1 y = 1 y = 1
Local max at ( 0 , 0 ) (0, 0) ( 0 , 0 )
Increasing on ( โ โ , โ 2 ) (-\infty, -2) ( โ โ , โ 2 ) and ( โ 2 , 0 ) (-2, 0) ( โ 2 , 0 )
Decreasing on ( 0 , 2 ) (0, 2) ( 0 , 2 ) and ( 2 , โ ) (2, \infty) ( 2 , โ )
As x โ ยฑ โ x \to \pm\infty x โ ยฑ โ , g ( x ) โ 1 g(x) \to 1 g ( x ) โ 1
(
0
,
0
)
( 0 , 0 )
โ
โ โ
0
x
)
(
โ
e โ x
)
( 1 โ x ) < 0 (1-x) < 0 ( 1 โ x ) < 0
e 1 โ
โ
0.368
x
)
+
e โ x ( โ 1 )
( x โ 2 ) > 0 (x-2) > 0 ( x โ 2 ) > 0
e 2
2
โ
โ
0.271
2
)
2
Horizontal asymptote y = 0 y=0 y = 0 on the right
โ