Second Derivative Test

The second derivative test is used to find local maxima and minima by checking the second derivative at critical points. It applies only to functions that are twice differentiable.

For a real-valued function f(x) that is defined on a closed or bounded interval [a, b]. Let k be a point in this interval; the second derivative test is done as follows:

• Differentiate f(x) to get f'(x).
• Differentiate again to get f''(x).
• Set f'(x) = 0 to find the critical points.
• Calculate f''(x) at the points obtained.

By calculating the value of f''(k), we can arrive at the following three conditions:

Case 1: Local Minima

If f'(x) = 0 and k is the required point, then if f''(k) > 0, the point k is said to be the point of local minimum.

Case 2: Local Maxima

If f'(x) = 0 and k is the required point, then if f''(k) < 0, the point k is said to be the point of local maxima.

Case 3: Point of Inflection

If f'(k) = 0 and k is the required point, then if f''(x) = 0, the point k is said to be the point of inflection, and the function is said to have no point of local maxima or minima.

Example: Find local maxima or local minima of the function f(x) = x³ - 6x.

Solution:

Given f(x) = x³-6x

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 3x² - 6

Equate f'(x) to 0

3x² - 6 = 0
⇒ x = √2 or -√2

Now calculate f''(x)

f''(x) = 6x

At x = √2, f''(√2) = 6√2 > 0. This means that x = √2 is the point of local minima.
At x = -√2, f''(-√2) = -6√2 < 0. This means that x = -√2 is the point of local maxima.

Multivariable Second Derivative Test

Multivariable second derivative test is used in case the given function has two variables (say, x and y). This method makes use of partial differentiation to find the local maxima and local minima. According to this test:

• Let f be a function with two variables x and y.
• Take partial derivatives of function f with respect to each variable and set them equal to 0. Solve the equations obtained to get the values of x and y.

Now define:

D(x,y) = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2

Now based on the value of D(x, y), we have:

1. If D(x, y) < 0, then f has a saddle point at x and y

2. If D(x, y) = 0, then the test fails

3. If D(x, y) > 0, then

• If \frac{\partial ^ 2f}{\partial x^2} (x,y) > 0, then (x, y) is the point of local minima
• If \frac{\partial ^ 2f}{\partial x^2} (x,y) < 0, then (x, y) is the point of local maxima

First vs Second Derivative Test

The first derivative test and the second derivative test are both used to find local maxima and minima, but there is a certain difference between the two. The following table lists the key differences between both the tests.

Limitations

• Indeterminate Results (Zero Second Derivative): If f′′(c) = 0 at a critical point c, the test is inconclusive.
• Requires Continuity: If the second derivative is discontinuous or does not exist, the test cannot be applied.
• Only Applies to Local Extrema: Useful only for determining local maxima or minima. It does not help in determining the global behavior of the function or identifying inflection points.

Solved Examples

Example 1: Find the point of local maxima and local minima of the function x³ - 12x using the second derivative test.

Solution:

Given f(x) = x³-12x

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 3x² - 12

Equate f'(x) to 0

3x² - 12 = 0
⇒ 3x² = 12
⇒ x = 2 or -2

Now calculate f''(x)

f''(x) = 6x

At x = 2, f''(2) = 12 > 0. This means that x = 2 is the point of local minima.
At x = -2, f''(-2) = -12 < 0. This means that x = -2 is the point of local maxima.

Example 2: Find the point of local maxima and local minima of the function x³ - x² - 5x using the second derivative test.

Solution:

Given f(x) = x³-x²- 5x

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 3x² - 2x - 5

Equate f'(x) to 0

3x² - 2x - 5 = 0
⇒ x = \frac{2\pm \sqrt{4-4(3)(-5)}}{2(3)}\\\implies x = \frac{2\pm \sqrt{64}}{6}\\\implies x = \frac{5}{3}~or~-1

Now calculate f''(x)

f''(x) = 6x - 2

At x = 5/3, f''(5/3) = 8 > 0. This means that x = 5/3 is the point of local minima.
At x = -1, f''(-1) = -8 < 0. This means that x = -1 is the point of local maxima.

Example 3: Find the point of local maxima and local minima of the function x - x⁴ using the second derivative test.

Solution:

Given f(x) = x⁴-x²

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 4x³ - 2x

Equate f'(x) to 0

4x³ - 2x = 0
⇒ 4x³ = 2x
⇒ 2x² = 1

x = 1/√2 or -1√2

Now calculate f''(x)

f''(x) = 12x² - 2

At x = 1/√2, f''(1/√2) = 4 > 0. This means that x = 1/√2 is the point of local minima.
At x = -1/√2, f''(-1/√2) = -8 < 0. This means that x = -1/√2 is the point of local maxima.

Example 4: A bus is moving along the curve 2y = 2x + 10. A man standing at point (3,5) wants to find the nearest distance between him and the bus. Calculate the nearest distance.

Solution:

Given 2y = 2x² + 10

⇒ y = x² + 10
f(x) = x² + 5

We need to calculate the nearest distance which means minimum distance. Let the distance be minimum when bus is at a point (x, y)

Distance of point (3, 5) from the point (x, y) = D = √(x-3)²+(y-5)²

D² = (x-3)² + (y-5)²

Substituting the value of y = x² + 5

D² = (x-3)² + (x²)² . . . (1)

D is minimum when D² is minimum. Thus we will calculate D'

D' = 2(x-3) + 4x³
⇒ D' = 2x - 6 + 4x³
⇒ D' = (x−1)(4x²+4x+6) = 2(x−1)(2x²+2x+3)

Equating D' = 0

x - 1 = 0 or 2x² + 2x + 3 = 0
⇒x = 1 OR
x = \frac{-2\pm \sqrt{4-4(2)(3)}}{2(2)}
\implies x = \frac{-2\pm \sqrt{-20}}{4}

This is not possible as x does not assume real value

Thus x = 1

Now D'' = 2+12x²

At x =1, D'' = 14 > 0. Thus x = 1 is the point of local minima.

Thus minimum distance is at x = 1 which is calculated by substituting x = 1 in equation (1)

D² = 4 + 1 = 5
⇒ D = √5

Thus the minimum distance is √5

Example 5: Calculate the maximum height of a cricket ball if its height is given by the function h(x) = 2x - 12x + 1 using the second derivative test.

Solution:

Given h(x) = 2x³- 12x²+1

We need to calculate the maximum height.

In order to calculate the maximum height, differentiate h(x) w.r.t x

h'(x) = 6x²- 24x

Equate h'(x) to 0
6x²- 24x = 0
⇒ 6x(x-4) = 0
⇒ x = 0 or 4

Now calculate f''(x)

h''(x) = 12x - 24

At x = 0, h''(0) = -24 < 0. This means that x = 1/√2 is the point of local maxima.
At x = 4, h''(4) = 24 > 0. This means that x = -1/√2 is the point of local minima.

Thus the height is maximum at a x = 0.

Maximum height = h(0) = 2(0)³-12(0) + 1 = 1 unit

Example 6: Find the point of local maxima and local minima of the function x⁴ - 12x³ using the second derivative test.

Solution:

Given f(x) = x⁴ - 12x³

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 4x³ - 36x²

Equate f'(x) to 0

4x³ - 36x² = 0
⇒ 4x²(x-9) = 0
⇒ x = 0 or 9

Now calculate f''(x)

f''(x) = 12x² - 72x

At x = 0, f''(0) = 0 > 0. This means that x = 0 is the point of inflection.
At x = 9, f''(9) = 12(9)²-72(9) = 324 > 0. This means that x = 9 is the point of local minima.

Example 7: Find the point of local maxima or local minima of the function x² - 1 using the second derivative test.

Solution:

Given f(x) = x² - 1

In order to calculate the local maxima and local minima, differentiate f(x) w.r.t x

f'(x) = 2x

Equate f'(x) to 0

2x = 0
⇒ x = 0

Now calculate f''(x)

f''(x) = 2

At x = 0, f''(0) = 2 > 0. This means that x = 0 is the point of local minima.

Example 8: Use the Second Derivative Test to determine the nature of the critical point of f(x) = x^2 e^(-x).

Solution:

First derivative: f'(x) = 2xe^(-x) - x^2e^(-x) = xe^(-x)(2 - x)

Critical points: x = 0 or x = 2

Second derivative: f''(x) = e^(-x)(2 - x) - xe^(-x)(2 - x) - xe^(-x) = e^(-x)(2 - 4x + x^2)

Evaluate f''(x) at critical points:

f''(0) = 2 > 0, so x = 0 is a local minimum
f''(2) = -2e^(-2) < 0, so x = 2 is a local maximum

Practice Problems

1. Use the Second Derivative Test to find and classify the critical points of f(x) = x³ - 3x² - 9x + 5.

2. Determine the nature of the critical points for the function f(x) = x⁴ - 8x² + 16.

3. Apply the Second Derivative Test to classify the critical points of f(x) = x^1/3 (x - 3)²

4. Find and classify the critical points of f(x) = xe^⁻ˣ² using the Second Derivative Test.

5. Use the Second Derivative Test to analyze the critical points of f(x) = ln(x² + 1) + 2x.

6. Determine the nature of the critical points for the function f(x) = x^2/3 - 2x^1/3.

7. Apply the Second Derivative Test to the function f(x) = x² / (x² + 1) and classify its critical points.

8. Find and classify the critical points of f(x) = x³⁵ - 5x²⁴ + 5x using the Second Derivative Test.

9. Use the Second Derivative Test to analyze the critical points of f(x) = sin(x) + cos(x) on the interval [0, 2π].