The Second Derivative Test: Videos & Practice Problems

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. Specifically, if \( f''(c) < 0 \) at a critical point \( c \), the function is concave down, indicating a local maximum. Conversely, if \( f''(c) > 0 \), the function is concave up, suggesting a local minimum. If \( f''(c) = 0 \), the test is inconclusive, and one must revert to the first derivative test.

To illustrate this process, consider the function \( f(x) = x^3 - 3x^2 + 4 \). The first step is to compute the first derivative, \( f'(x) \), which is found using the power rule:

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

Setting \( f'(x) = 0 \) gives critical points at \( x = 0 \) and \( x = 2 \). Next, we calculate the second derivative:

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

Now, we evaluate the second derivative at the critical points:

For \( x = 0 \):

\( f''(0) = 6(0) - 6 = -6 \) (negative, indicating a local maximum)

For \( x = 2 \):

\( f''(2) = 6(2) - 6 = 6 \) (positive, indicating a local minimum)

Thus, the local extrema are identified: a local maximum at \( x = 0 \) and a local minimum at \( x = 2 \). To find the corresponding function values, substitute these critical points back into the original function:

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

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

In summary, the local maximum occurs at \( (0, 4) \) and the local minimum at \( (2, 2) \). This method of using the second derivative test is an efficient way to determine the nature of critical points in a function.

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 is essential for identifying 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:

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

Further factoring reveals that:

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

Setting the first derivative equal to zero to find critical points, we have:

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

This leads to the critical point \( x = 1 \). Since \( f(x) \) is a polynomial, the first derivative exists everywhere, allowing us to proceed with the second derivative test.

Next, we compute the second derivative:

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

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

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

Since the second derivative is zero, the second derivative test is inconclusive. We must then use the first derivative test to determine the nature of the critical point.

To apply the first derivative test, we analyze the sign of \( f'(x) \) around the critical point. We create a sign chart with intervals \( (-\infty, 1) \) and \( (1, \infty) \). Choosing test points, we select \( x = 0 \) for the first interval and \( x = 2 \) for the second interval:

For \( x = 0 \):

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

For \( x = 2 \):

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

Since \( f'(x) \) changes from negative to positive at \( x = 1 \), this indicates that \( x = 1 \) is a local minimum.

To find the minimum value, we evaluate the original function at this critical point:

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

Thus, 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 found a local minimum at \( x = 1 \) with a minimum value of \( 0 \).