5.7 Using the Second Derivative Test to Determine Extrema

The Second Derivative Test is a powerful tool in calculus that can be used to determine the nature of extrema (maxima and minima) of a function. It is based on the fact that the sign of the second derivative of a function can indicate whether a critical point is a maximum, minimum, or neither. Here are the steps to use the Second Derivative Test: 1. Find the critical points of the function by setting the first derivative equal to zero and solving for x. 2. Find the second derivative of the function. 3. Evaluate the second derivative at each critical point. 4. If the second derivative is positive at a critical point, the function has a local minimum at that point. 5. If the second derivative is negative at a critical point, the function has a local maximum at that point. 6. If the second derivative is zero at a critical point, the test is inconclusive and further analysis is needed. It is important to note that the Second Derivative Test can only be applied to a function that is twice differentiable, and that it only gives information about the extrema at a single point, not on an interval.

Example 1: Consider the function f(x) = x^2. The first derivative of this function is f'(x) = 2x, and the second derivative is f''(x) = 2. The critical point is x = 0. Since the second derivative is positive at x = 0, the function has a local minimum at x = 0. Example 2: Consider the function f(x) = -x^2. The first derivative of this function is f'(x) = -2x, and the second derivative is f''(x) = -2. The critical point is x = 0. Since the second derivative is negative at x = 0, the function has a local maximum at x = 0. Example 3: Consider the function f(x) = x^3. The first derivative of this function is f'(x) = 3x^2, and the second derivative is f''(x) = 6x. The critical point is x = 0. Since the second derivative is zero at x = 0, the test is inconclusive, and further analysis is needed to determine the nature of the extrema. Example 4: Consider the function f(x) = x^4 - 6x^2. The first derivative of this function is f'(x) = 4x^3 - 12x, and the second derivative is f''(x) = 12x^2 - 12. The critical points are x = +/- sqrt(3). To find the extrema, we need to find the sign of the second derivative at x = +/- sqrt(3). Since 12x^2 - 12 is positive at x = +/- sqrt(3), the function has a local minimum at x = +/- sqrt(3) Example 5: Consider the function f(x) = x^4 +6x^2. The first derivative of this function is f'(x) = 4x^3 +12x, and the second derivative is f''(x) = 12x^2 + 12. The critical points are x = 0. To find the extrema, we need to find the sign of the second derivative at x = 0. Since 12x^2 + 12 is positive at x = 0, the function has a local minimum at x = 0.

Example 6: Consider the function f(x) = sin(x). The first derivative of this function is f'(x) = cos(x), and the second derivative is f''(x) = -sin(x). The critical points are x = kpi, where k is any integer. To find the extrema, we need to find the sign of the second derivative at x = kpi. Since -sin(x) is negative at x = kpi, the function has a local maximum at x = kpi.

Example 7: Consider the function f(x) = e^x. The first derivative of this function is f'(x) = e^x, and the second derivative is f''(x) = e^x. The critical point is x = 0. Since the second derivative is positive at x = 0, the function has a local minimum at x = 0.

Example 8: Consider the function f(x) = x^2 + 2x + 3. The first derivative of this function is f'(x) = 2x + 2, and the second derivative is f''(x) = 2. The critical point is x = -1. Since the second derivative is positive at x = -1, the function has a local minimum at x = -1.

Example 9: Consider the function f(x) = x^3 - 6x^2 + 11x - 6. The first derivative of this function is f'(x) = 3x^2 - 12x + 11, and the second derivative is f''(x) = 6x - 12. The critical points are x = 1 and x = 2/3. To find the extrema, we need to find the sign of the second derivative at x = 1 and x = 2/3. Since the second derivative is negative at x = 1, the function has a local maximum at x = 1. And since the second derivative is positive at x = 2/3, the function has a local minimum at x = 2/3.

Example 10: Consider the function f(x) = x^4 - 4x^3 + 4x^2. The first derivative of this function is f'(x) = 4x^3 - 12x^2 + 8x, and the second derivative is f''(x) = 12x^2 - 24x + 8. The critical points are x = 0 and x = 1. To find the extrema, we need to find the sign of the second derivative at x = 0 and x = 1. Since the second derivative is positive at x = 0, the function has a local minimum at x = 0. And since the second derivative is negative at x = 1, the function has a local maximum at x = 1.

In conclusion, the Second Derivative Test is a powerful tool in calculus that can be used to determine the nature of extrema (maxima and minima) of a function. By finding the critical points, evaluating the second derivative, and determining its sign, we can determine whether a critical point is a maximum, minimum, or neither. However, it is important to remember that the Second Derivative Test can only be applied to twice differentiable functions, and it only gives information about the extrema at a single point, not on an interval.