The Second Derivative Test: Videos & Practice Problems

The second derivative test is a method for finding local extrema of a function by analyzing the sign of the second derivative. If the second derivative is negative at a critical point, the function has a local maximum; if positive, it has a local minimum. If the second derivative equals zero, the first derivative test must be used. For example, for the function $f\left(x\right)=x^3-3x^2+4$, critical points at $x=0$ (local max) and $x=2$ (local min) are identified using this test.

The Second Derivative Test: Finding Local Extrema

To find local extrema of a function, the second derivative test is a powerful method that utilizes the sign of the second derivative. This approach complements the first derivative test, which identifies critical points where the first derivative is zero or undefined. The second derivative, denoted as \( f''(x) \), provides insight into the concavity of the function, which is crucial for determining whether a critical point is a local maximum or minimum.

When applying the second derivative test, follow these steps:

1. **Identify Critical Points**: Start by finding the first derivative \( f'(x) \) of the function and set it equal to zero to locate critical points. For example, if \( f(x) = x^3 - 3x^2 + 4 \), the first derivative is calculated as:

\[ f'(x) = 3x^2 - 6x = 3x(x - 2) \]

Setting \( f'(x) = 0 \) gives critical points at \( x = 0 \) and \( x = 2 \).

2. **Evaluate the Second Derivative**: Next, compute the second derivative \( f''(x) \). For the function above, the second derivative is:

\[ f''(x) = 6x - 6 = 6(x - 1) \]

3. **Determine the Sign of the Second Derivative**: Plug the critical points into the second derivative to assess its sign:

- For \( x = 0 \):

\[ f''(0) = 6(0) - 6 = -6 \] (negative)

This indicates that the function is concave down at \( x = 0 \), confirming a local maximum.

- For \( x = 2 \):

\[ f''(2) = 6(2) - 6 = 6 \] (positive)

This indicates that the function is concave up at \( x = 2 \), confirming a local minimum.

4. **Conclusion**: From the evaluations, we conclude that there is a local maximum at \( x = 0 \) and a local minimum at \( x = 2 \). To find the actual maximum and minimum values, substitute these critical points back into the original function:

- For \( x = 0 \): \( f(0) = 0^3 - 3(0)^2 + 4 = 4 \) (local maximum value)

- For \( x = 2 \): \( f(2) = 2^3 - 3(2)^2 + 4 = -2 \) (local minimum value)

In summary, the second derivative test is an effective method for identifying local extrema by analyzing the concavity of the function at critical points. Remember, if the second derivative equals zero, further analysis using the first derivative test is necessary.

Use the second derivative test to find the local extrema of the given function.
$g(x)=x^3-6x^2+9x+2$

• A. Local max of $6$ at $x=1$, local min of 2 at $x=3$
• B. Local min of $6$ at $x=1$, local max of 2 at $x=3$
• C. Local max of $6$ at $x=-1$ and $x=1$, local min of $-6$ at $x=-3$ and $x=3$
• D. No local extrema

Answer: (A)

Use the second derivative test to find the local extrema of the given function.
$f\left(x\right)=\frac{x^2-4}{x^2+1}$

• A. Local max of $10$ at $x=0$
• B. Local min of $-4$ at $x=0$
• C. Local min of $-4$ at $x=-2$ and Local max of $10$ at $x=2$
• D. No local extrema

Answer: (B)

Use the second derivative test to find the local extrema of the given function.
$f\left(x\right)=4\sin x\cos x$; $0$ < $x$ < $\pi$

• A. Local min of $-2$ at $x=\frac{\pi}{2}$, local max of $8$ at $x=\pi$
• B. Local max of $2$ at $x=\frac{\pi}{2}$, local min of $-2$ at $x=\pi$
• C. Local min of $-8$ at $x=\frac{\pi}{4}$, local max of $8$ at $x=\frac{3\pi}{4}$
• D. Local max of $2$ at $x=\frac{\pi}{4}$, local min of $-2$ at $x=\frac{3\pi}{4}$

Answer: (D)

The Second Derivative Test: Finding Local Extrema Example 1

To find the local extrema of the function \( f(x) = x^4 - 4x^3 + 6x^2 - 4x + 1 \), we start by calculating the first derivative, which helps identify critical points. Using the power rule, the first derivative is given by:

\( f'(x) = 4x^3 - 12x^2 + 12x - 4 \)

Factoring out a common factor of 4, we can simplify this to:

\( f'(x) = 4(x^3 - 3x^2 + 3x - 1) \)

This expression can be factored further to:

\( f'(x) = 4(x - 1)^3 \)

Setting the first derivative equal to zero, we find:

\( 4(x - 1)^3 = 0 \implies x - 1 = 0 \implies x = 1 \)

Next, we apply the second derivative test to determine the nature of this critical point. The second derivative is calculated as follows:

\( f''(x) = 12(x - 1)^2 \)

Evaluating the second derivative at \( x = 1 \) gives:

\( f''(1) = 12(1 - 1)^2 = 12 \cdot 0 = 0 \)

Since the second derivative is zero, the second derivative test is inconclusive. Therefore, we must use the first derivative test to analyze the behavior of the function around the critical point.

We create a sign chart for \( f'(x) \) by testing values in the intervals \( (-\infty, 1) \) and \( (1, \infty) \). Choosing \( x = 0 \) for the first interval:

\( f'(0) = 4(0 - 1)^3 = 4(-1)^3 = -4 \) (negative)

Choosing \( x = 2 \) for the second interval:

\( f'(2) = 4(2 - 1)^3 = 4(1)^3 = 4 \) (positive)

From this analysis, we see that \( f'(x) \) changes from negative to positive at \( x = 1 \), indicating that the function is decreasing before \( x = 1 \) and increasing after \( x = 1 \). Thus, \( x = 1 \) is a local minimum.

To find the value of the local minimum, we substitute \( x = 1 \) back into the original function:

\( f(1) = 1^4 - 4(1^3) + 6(1^2) - 4(1) + 1 = 1 - 4 + 6 - 4 + 1 = 0 \)

Therefore, the function has a local minimum at \( x = 1 \) with a minimum value of \( 0 \).

In summary, when using the second derivative test, if the result is zero, it is necessary to revert to the first derivative test to determine the nature of the critical point. In this case, we confirmed that \( x = 1 \) is a local minimum with a value of \( 0 \).

Here’s what students ask on this topic:

What is the second derivative test in calculus?

The second derivative test is a method used in calculus to determine the local extrema (maximum or minimum) of a function. It involves taking the second derivative of the function and analyzing its sign at critical points where the first derivative is zero. If the second derivative is negative at a critical point, the function has a local maximum there. If the second derivative is positive, the function has a local minimum. If the second derivative is zero, the test is inconclusive, and the first derivative test must be used instead.

How do you find the local extrema of a function using the second derivative test?

To find the local extrema of a function using the second derivative test, follow these steps: 1) Find the first derivative of the function and determine the critical points by setting the first derivative equal to zero. 2) Compute the second derivative of the function. 3) Evaluate the second derivative at each critical point. 4) Analyze the sign of the second derivative at each critical point: if it is negative, the function has a local maximum; if it is positive, the function has a local minimum. If the second derivative is zero, the test is inconclusive, and the first derivative test should be used.

What does it mean if the second derivative is zero at a critical point?

If the second derivative is zero at a critical point, the second derivative test is inconclusive. This means that the test does not provide any information about whether the critical point is a local maximum, local minimum, or neither. In such cases, you need to use the first derivative test or other methods to determine the nature of the critical point.

Can the second derivative test be used if the first derivative does not exist at a critical point?

No, the second derivative test cannot be used if the first derivative does not exist at a critical point. The second derivative test requires that the first derivative be zero at the critical point. If the first derivative does not exist, you must use the first derivative test or other methods to analyze the critical point.

How do you apply the second derivative test to the function f(x) = x^3 - 3x^2 + 4?

To apply the second derivative test to the function f(x) = x^3 - 3x^2 + 4, follow these steps: 1) Find the first derivative: f'(x) = 3x^2 - 6x. 2) Determine the critical points by setting the first derivative equal to zero: 3x^2 - 6x = 0, which gives x = 0 and x = 2. 3) Find the second derivative: f''(x) = 6x - 6. 4) Evaluate the second derivative at the critical points: f''(0) = -6 (negative, indicating a local maximum at x = 0) and f''(2) = 6 (positive, indicating a local minimum at x = 2).