MathLab with Mr. Ahmed

MathLab with Mr. Ahmed is a math learning website by Ahmed Amasha, a mathematics teacher teaching the American curriculum. Explore Algebra 2, Precalculus, Calculus, and SAT and PSAT Math Prep through lessons, focused practice and printable resources.

MathLabwith Mr. Ahmed
Applications of Differentiation

3.4 Concavity and the Second Derivative Test

Focus on concavity, inflection points, second derivative test.

Free preview Explore the lesson before creating an account.

Play games for this section

Before you solve

Ideas in this lesson

01ConcavityWorked steps · five questions02Inflection pointsWorked steps · five questions03Second Derivative TestWorked steps · five questions
Idea 1 of 3

Concavity

Positive f′′f'' indicates increasing tangent slopes and concave-up behavior; negative f′′f'' indicates concave-down behavior.

Worked example 1

For f(x)=x3−6x2f(x) = x^{3} - 6x^{2}, find the xx-coordinate of its inflection point.
  1. f′′(x)=6(x−2)f''(x) = 6(x - 2)
  2. f′′<0f'' < 0 for x<2x < 2; f′′>0f'' > 0 for x>2x > 2
  3. Inflection x=2x = 2

Answer: 22

Worked example 2

For f(x)=x3−9x2f(x) = x^{3} - 9x^{2}, find the inflection-point yy-coordinate.
  1. f′′(x)=6(x−3)f''(x) = 6(x - 3)
  2. f′′<0f'' < 0 for x<3x < 3; f′′>0f'' > 0 for x>3x > 3
  3. f(3)=33−9(3)2=−54f(3) = 3^{3} - 9(3)^{2} = -54

Answer: −54-54

Idea 2 of 3

Inflection points

Concavity must change at an inflection point. Solving f′′=0f'' = 0 only identifies candidates.

Worked example 1

For f(x)=x3−6x2f(x) = x^{3} - 6x^{2}, find the xx-coordinate of its inflection point.
  1. f′′(x)=6(x−2)f''(x) = 6(x - 2)
  2. f′′<0f'' < 0 for x<2x < 2; f′′>0f'' > 0 for x>2x > 2
  3. Inflection x=2x = 2

Answer: 22

Worked example 2

For f(x)=x3−9x2f(x) = x^{3} - 9x^{2}, find the inflection-point yy-coordinate.
  1. f′′(x)=6(x−3)f''(x) = 6(x - 3)
  2. f′′<0f'' < 0 for x<3x < 3; f′′>0f'' > 0 for x>3x > 3
  3. f(3)=33−9(3)2=−54f(3) = 3^{3} - 9(3)^{2} = -54

Answer: −54-54

Idea 3 of 3

Second Derivative Test

At a critical point, positive f′′f'' gives a minimum and negative f′′f'' gives a maximum. When f′′=0f'' = 0, the test is inconclusive.

Worked example 1

Use the Second Derivative Test for f(x)=(x−2)2f(x) = (x - 2)^{2} at its critical point x=2x = 2.
  1. f′(2)=0f'(2) = 0
  2. f′′(2)=2>0f''(2) = 2 > 0
  3. f′(2)=0,f′′(2)=2  ⟹  f'(2) = 0, f''(2) = 2 \implies Local minimum

Answer: Local minimum

Worked example 2

Use the Second Derivative Test for f(x)=−(x−3)2f(x) = -(x - 3)^{2} at its critical point x=3x = 3.
  1. f′(3)=0f'(3) = 0
  2. f′′(3)=−2<0f''(3) = -2 < 0
  3. f′(3)=0,f′′(3)=−2  ⟹  f'(3) = 0, f''(3) = -2 \implies Local maximum

Answer: Local maximum

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §3.4,p3.4, p. 187187. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.