First Derivative Test

The First Derivative Test is a crucial concept in calculus used to determine the behavior of a function, particularly in identifying local maxima, minima, and points of inflection. This method helps analyze how a function changes its slope and whether it reaches its highest or lowest point within a particular interval. In this blog, we will explore the definition, applications, and relationship between the First Derivative Test and Second Derivative Test. We will also look at a first derivative test example to deepen our understanding.

1.0What is the First Derivative Test

The First Derivative Test involves using the derivative of a function, denoted as f'(x), to determine whether a given point is a local maximum, minimum, or neither. By analyzing the sign changes of f'(x), we can ascertain the nature of critical points, which are the values of x where f'(x) = 0 or f'(x) is undefined.

Steps Involved in the First Derivative Test

1. Find the First Derivative f'(x): Calculate the derivative of the function.
2. Identify Critical Points: Solve f'(x) = 0 or find where f'(x) is undefined. These points are potential candidates for local maxima or minima.
3. Determine the Sign of f'(x) on Either Side of Each Critical Point: Choose test points in the intervals formed by the critical points and analyze whether f'(x) is positive or negative.
4. Conclude the Nature of Each Critical Point:

• If f'(x) changes from positive to negative, the point is a local maximum.
• If f'(x) changes from negative to positive, the point is a local minimum.
• If f'(x) does not change sign, the point is neither a maximum nor a minimum.

2.0Relationship Between First Derivative Test and Second Derivative Test

While the First Derivative Test relies on analyzing the slope changes, the Second Derivative Test involves the second derivative f''(x). The second derivative provides information about the concavity of the function:

• If f''(x) > 0 at a critical point, the function is concave up, and the point is a local minimum.
• If f''(x) < 0 at a critical point, the function is concave down, and the point is a local maximum.
• If f''(x) = 0, the test is inconclusive, and we revert to the First Derivative Test for further analysis.

By using both the first derivative test and second derivative test, we can comprehensively understand the function's behavior at critical points and their surroundings.

3.0First Derivative Test for Maxima and Minima

The First Derivative Test is particularly useful for identifying maxima and minima in a function without needing to evaluate the second derivative. It provides a straightforward method to analyze the behavior of a function around its critical points.

Key Points to Remember

Local Maximum: f'(x) changes from positive to negative at a critical point.

Local Minimum: f'(x) changes from negative to positive at a critical point.

No Maximum or Minimum: f'(x) does not change sign at a critical point.

4.0Solved Example on First Derivative Test

Example 1: Consider the function $f(x)=x^3-3x^2+2.$

Solution: 

Find the First Derivative:

$f^{\prime }(x)=3x^2-6x$

Identify Critical Points:

Solve f'(x) = 0:

• $3x^2-6x=0$
• $3x(x-2)=0$

The critical points are x = 0 and x = 2.

Determine the Sign of f'(x):

• For x < 0, let’s choose x = –1:

$f^{\prime }(-1)=3(-1{)}^2-6(-1)=3+6=9\text{ (Positive) }$

• For 0 < x < 2, let’s choose x = 1:

$f^{\prime }(1)=3(1{)}^2-6(1)=3-6=-3\text{ (Negative) }$

• For x > 2, let’s choose x = 3:

$f^{\prime }(3)=3(3{)}^2-6(3)=27-18=9\text{ (Positive) }$

Conclude the Nature of Each Critical Point:

• At x = 0, f'(x) changes from positive to negative, indicating a local maximum.
• At x = 2, f'(x) changes from negative to positive, indicating a local minimum.

Thus, the First Derivative Test helps identify the local maximum at x = 0 and the local minimum at x = 2 for the function $f(x)=x^3-3x^2+2.$

Example 2: Find its local maxima and minima of the function $f(x)=2x^3-3x^2-12x+5$  using First Derivative Test.

Solution:

Find the First Derivative

Calculate the first derivative f'(x) of the function f(x):

$f^{\prime }(x)=6x^2-6x-12$

Determine the Critical Points

To find critical points, set f'(x) = 0 and solve for x:

$6x^2-6x-12=0$

Divide through by 6:

$x^2-x-2=0$

Factor the quadratic equation:

(x-2)(x+1)=0

The critical points are x = 2 and x = –1.

Determine the Sign of f'(x) on Either Side of Each Critical Point

Next, we choose test points in the intervals defined by the critical points: (–∞, –1), (–1, 2), and (2, ∞).

1. Interval (–∞, –1): 

Choose x = –2:

$\begin{array}{rl}& f^{\prime }(-2)=6(-2{)}^2-6(-2)-12=24+12-12=24\\ & \text{ (Positive) }\end{array}$

2. Interval (–1, 2):  

Choose x = 0:

$f^{\prime }(0)=6(0{)}^2-6(0)-12=-12\text{ (Negative) }$

3. Interval (2, ∞):  

Choose x = 3:

$f^{\prime }(3)=6(3{)}^2-6(3)-12=54-18-12=24\text{ (Positive) }$

Conclude the Nature of Each Critical Point

Now, based on the sign changes of f'(x):

• At x = –1, f'(x) changes from positive to negative. This indicates a local maximum at x = –1.
• At x = 2, f'(x) changes from negative to positive. This indicates a local minimum at x = 2.

Determine the Function Values at the Critical Points

Calculate f(x) at the critical points to find the maximum and minimum values:

1. At x = –1:

$f(-1)=2(-1{)}^3-3(-1{)}^2-12(-1)+5$

= –2 –3 + 12 + 5 = 12

So, the local maximum is (–1, 12).

2. At x = 2:

$f(2)=2(2{)}^3-3(2{)}^2-12(2)+5$

= 16 – 12 –24 + 5 = –15

So, the local minimum is (2, –15).

Example 3: Finding Maxima and Minima of $f(x)=x^4-8x^2+16$

Solution:

Find the First Derivative

$f^{\prime }(x)=4x^3-16x$

Determine the Critical Points**

Set f'(x) = 0:

• $4x^3-16x=0$
• $4x\left(x^2-4\right)=0$
• $4x(x-2)(x+2)=0$

The critical points are x = 0, x = 2, x = –2.

Determine the Sign of f'(x) in Each Interval

Choose test points in the intervals (–∞, –2), (–2, 0), (0, 2), and (2, ∞).

• Interval $(-\infty ,-2)$ :
• Choose x = –3:$f^{\prime }(-3)=4(-3{)}^3-16(-3)=-108+48=-60\text{ (Negative) }$
• Interval (–2, 0): 

Choose x = –1:

$f^{\prime }(-1)=4(-1{)}^3-16(-1)$ = –4 + 16 = 12 (Positive)

• Interval (0, 2):  

  Choose x = 1:

$f^{\prime }(1)=4(1{)}^3-16(1)$

= 4 – 16 = –12 (Negative)

• Interval (2, ∞):  

  Choose x = 3:

$f^{\prime }(3)=4(3{)}^3$ - 16(3)

= 108 – 48

= 60 (Positive)

Determine the Nature of Critical Points

At x = –2, f'(x) changes from negative to positive. So, x = –2 is a local minimum.

At x = 0, f'(x) changes from positive to negative. So, x = 0 is a local maximum.

At x = 2, f'(x) changes from negative to positive. So, x = 2 is a local minimum.

Determine the Function Values at Critical Points

$f(-2)=(-2{)}^4-8(-2{)}^2+16=16-32+16=0$

$f(0)=(0{)}^4-8(0{)}^2+16=16$

$f(2)=(2{)}^4-8(2{)}^2+16=16-32+16=0$

Example 4: Maxima and Minima of $f(x)=3x^5-5x^3$

Solution:

Find the First Derivative

$f^{\prime }(x)=15x^4-15x^2$

Determine the Critical Points

Set f'(x) = 0:

• $15x^4-15x^2=0$
• $15x^2\left(x^2-1\right)=0$
• $15x^2(x-1)(x+1)=0$

The critical points are x = 0, x = 1, x = –1.

Determine the Sign of f'(x) in Each Interval

Choose test points in the intervals (–∞, –1), (–1, 0), (0, 1), and (1, ∞). 

• Interval (–∞, –1): 

Choose x = –2:

$f^{\prime }(-2)=15(-2{)}^4-15(-2{)}^2$

= 180 (Positive)

• Interval (–1, 0):  

Choose $x=-\frac{1}{2}$ :

$f^{\prime }\left(-\frac{1}{2}\right)=15{\left(-\frac{1}{2}\right)}^4-15{\left(-\frac{1}{2}\right)}^2$

= 0.9375 – 3.75 

= –2.8125 (Negative)

• Interval (0, 1):  

Choose $x=\frac{1}{2}$ :

$f^{\prime }\left(\frac{1}{2}\right)=15{\left(\frac{1}{2}\right)}^4-15{\left(\frac{1}{2}\right)}^2=0.9375-3.75=-2.8125\text{ (Negative) }$

• Interval (1, ∞):   

Choose x = 2:

$f^{\prime }(2)=15(2{)}^4-15(2{)}^2$

= 240 – 60

= 180 (Positive)

Determine the Nature of Critical Points

At x = –1, f'(x) changes from positive to negative. So, x = –1 is a local maximum.

At x = 0, f'(x) does not change sign (negative on both sides), so x = 0 is neither a maximum nor a minimum.

At x = 1, f'(x) changes from negative to positive. So, x = 1 is a local minimum.

Determine the Function Values at Critical Points

$f(-1)=3(-1{)}^5-5(-1{)}^3=-3+5=2$

$f(0)=3(0{)}^5-5(0{)}^3=0$

$f(1)=3(1{)}^5-5(1{)}^3=3-5=-2$

5.0Practice Questions for First Derivative Test

1. Find the local maxima and minima of the function: $f(x)=x^3-3x^2-9x+7$
2. Given the function: $f(x)=x^4-8x^3+18x^2$Use the first derivative test to find the local maxima and minima. Determine the coordinates of these points.
3. For the function: $f(x)=3x^5-5x^4+2$, Use the first derivative test to find the critical points and evaluate whether they correspond to local maxima or minima.
4. Find the local maxima and minima of the function: $f(x)=x^2{e}^{-x}$. Apply the first derivative test to identify the nature of each critical point.
5. Determine the local maxima and minima of the function: $f(x)=\frac{x}{x^2+1}$. Use the first derivative test to find and classify the critical points.