#Second Derivative Test

#Table of Contents

First and Second Derivatives at Key Points
What is the Second Derivative Test?
Points with a Second Derivative of Zero
Worked Examples
Practice Questions
Glossary
Summary and Key Takeaways

#First and Second Derivatives at Key Points

To be able to classify key points on the graph of a function, it is important that you are confident with the properties of the first and second derivatives at these points.

Type of Point	First Derivative	Second Derivative
Local minimum	Zero	Positive or zero
Local maximum	Zero	Negative or zero
Point of inflection (critical)	Zero	Zero
Point of inflection (non-critical)	Non-zero	Zero

Key Concept

It is essential to understand the behavior of first and second derivatives at these key points for effective classification.

#What is the Second Derivative Test?

The second derivative can be used to determine if a critical point is a local minimum or maximum.

The second derivative test states that:

If $f'(a)=0$ and $f''(a)>0$ , then $f(x)$ has a local minimum at $x=a$ .
If $f'(a)=0$ and $f''(a)<0$ , then $f(x)$ has a local maximum at $x=a$ .
If $f'(a)=0$ and $f''(a)=0$ , this test does not give any information; it could be any of a local minimum, local maximum, or point of inflection.

### Example: For

f(x) = x^3 - 3x^2 + 3

, find the local extrema and classify them using the second derivative test.

Solution:

Find the first derivative: $f'(x) = 3x^2 - 6x$ .
Set the first derivative to zero: $3x^2 - 6x = 0 \Rightarrow x(x-2) = 0 \Rightarrow x = 0$ or $x = 2$ .
Find the second derivative: $f''(x) = 6x - 6$ .
Evaluate the second derivative at the critical points:
- $f''(2) = 6 \times 2 - 6 = 6$ (positive, so $x=2$ is a local minimum).
- $f''(0) = 6 \times 0 - 6 = -6$ (negative, so $x=0$ is a local maximum).

Therefore, the local minimum is at $(2, -1)$ and the local maximum is at $(0, 3)$ .

#Points with a Second Derivative of Zero

All points of inflection have a second derivative of zero.
However, not all points with a second derivative of zero are points of inflection.
Local minimums or maximums can also have a second derivative of zero.

Key Concept

The second derivative test is only for determining if a critical point is a local minimum or maximum. If $f''(x)=0$ , further investigation is needed by applying the first derivative test.

#First Derivative Test:

Find the values of the first derivative:
- At an $x$ value slightly to the left of the critical point.
- At an $x$ value slightly to the right of the critical point.
If the first derivative changes (from left to right):
- From positive to negative, it is a local maximum.
- From negative to positive, it is a local minimum.
If the sign stays the same on both sides of the critical point, it is a point of inflection.

### Example: For

f(x) = x^3 - 3x^2 + 3

, determine if

(1, 1)

is a point of inflection.

Solution:

Find the second derivative and set it to zero: $f''(x) = 6x - 6 \Rightarrow 6x - 6 = 0 \Rightarrow x = 1$ .
Evaluate $y$ at $x=1$ : $f(1) = 1$ so the point is $(1, 1)$ .
Check the first derivative at $x=1$ : $f'(1) = 3 \times 1^2 - 6 \times 1 = -3$ (non-zero).

Therefore, $(1, 1)$ is a point of inflection.

#Worked Examples

#Example 1:

Determine the nature of the critical point for $g(x) = 3x^4 + 2x^6$ .

Solution:

Find the first derivative: $g'(x) = 12x^3 + 12x^5$ .
Set the first derivative to zero: $12x^3 + 12x^5 = 0 \Rightarrow 12x^3(1 + x^2) = 0 \Rightarrow x = 0$ .
Find the second derivative: $g''(x) = 36x^2 + 60x^4$ .
Evaluate the second derivative at the critical point: $g''(0) = 0$ .
Check the first derivative on either side of the point:
- $g'(-1) = -24$
- $g'(1) = 24$

The graph changes from decreasing to the left of $(0, 0)$ to increasing to the right of $(0, 0)$ . Therefore, $(0, 0)$ is a minimum point.

#Practice Questions

Practice Question

#Question 1:

Find the local extrema of the function $h(x) = x^4 - 4x^3 + 6x^2$ and classify them using the second derivative test.

Practice Question

#Question 2:

Determine whether the point $(2, \frac{32}{3})$ is a point of inflection for the function $k(x) = x^5 - 5x^4 + 10x^3$ .

#Glossary

Critical Point: A point on the graph where the first derivative is zero or undefined.
Local Minimum: A point where the function value is lower than all nearby points.
Local Maximum: A point where the function value is higher than all nearby points.
Point of Inflection: A point where the graph changes concavity.

#Summary and Key Takeaways

The second derivative test helps classify critical points as local minima or maxima.
If $f'(x) = 0$ and $f''(x) > 0$ , there is a local minimum.
If $f'(x) = 0$ and $f''(x) < 0$ , there is a local maximum.
If $f'(x) = 0$ and $f''(x) = 0$ , use the first derivative test for further investigation.
Points of inflection have a second derivative of zero but not all points with a second derivative of zero are points of inflection.