Solved Examples on Points of Inflection

Example 1: f(x)=x3.

Solution:

First derivative: f'(x)=3x2

Second derivative: f”(X)=6x

Find Potential Points of Inflection:

Set the second derivative equal to zero:

6x=0

x=0

Test for Sign Change:

for x<0 (e.g., x=-1):

f”(-1)=6(-1)=-6 (negative, concave down)

for x>0 (e.g., x=1)

f”(1)=6(1)=6 (positive , concave up)

Conclusion:

There is a point of inflection at x=0 because the concavity changes from down to up.

Example 2: g(x)=x4-4x3+6x2

Solution:

First derivative: g'(x)=4x3-12x2+12x

Second derivative: g”(x)=12x2-24x+12

Find Potential Points of Inflection:

Set the second derivative equal to zero:

12x2-24x+12=0

x2-2x+1=0

(x-1)2=0

x=1

Test for Sign Change:

for x<1 (e.g. , x=0)

g”(0)=12(0)2-24(0)+12=12 (positive, concave up)

for x>1 (e.g., x=2)

g”(2)=12(2)2-24(2)+12=12 (positive, concave up)

Conclusion:

There is no point of inflection at x=1 because the concavity does not change.

Example 3: h(x)=x3-3x2+3x-1

Solution:

First derivative: h'(x)=3x2-6x+3

Second derivative: h”(x)=6x-6

Find Potential Points of Inflection:

Set the second derivative equal to zero: 6x-6=0

x=1

Test for Sign Change:

for x<1 (e.g., x=0):

h”(0)=6(0)-6=-6 (negative, concave down)

for x>1(e.g., x=2):

h”(2)=6(2)-6=6 (positive , concave up)

Conclusion:

There is a point of inflection at x=1 because the concavity changes from down to up.

Example 4: k(x)=sin(x) on the interval [0,2π]

Solution:

First Derivative: k′(x)=cos(x)

Second Derivative: k′′(x)=−sin(x)

Find Potential Points of Inflection:

Set the second derivative equal to zero:

−sin(x)=0

So, x=0,π,2π.

Test for Sign Change:

  • For x=0:
    • For x<0 (e.g. x=-0.1): k”(-0.1)=-sin(0.1) ≈ 0.1(positive)
    • For x>0 (e.g., x=0.1): k”(0.1)=-sin(0.1) ≈ -0.1(negative)
  • For x= π:
    • For x<π(e.g. x=π-0.1): k”(π-0.1)=-sin(π-0.1) ≈0.1 (positive)
    • For x>π (e.g., x=π+0.1):k”(π+0.1)=-sin(π+0.1) ≈-0.1(negative)
  • For x=2π:
    • For x<2π (e.g., x=2π-0.1): k”(2π-0.1)=-sin(2π-0.1)≈ 0.1 (positive)
    • For x>2π (e.g., x=2π+0.1):k”(2π+0.1)=-sin(2π+0.1) ≈-0.1(negative)

Conclusion:

There are points of inflection at x=0,π,2π because the concavity changes at these points.

Concavity and Points of Inflection

Concavity and points of inflection are the key concepts and basic fundamentals of calculus and mathematical analysis. It provides an insight into how curves behave and the shape of the functions. Where concavity helps us to understand the curving of a function, determining whether it is concave upward or downward, the point of inflection determines the point where the concavity changes, i.e., where either curve transforms from concave upward to concave downward or concave to convex, and vice versa. These concepts are essential in various mathematical applications, including curve sketching, optimization problems , and the study of differential equations.

In this article, we’ll shed lights on the definitions, properties, and practical implications of concavity and points of inflection.

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Conclusion

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