Mastering The Second Derivative Test: How To Find Extrema And Concavity Of Functions

Second derivative test

if f'(x) = 0 and f”(x) > 0 then minimum; if f'(x) = 0 and f”(x) < 0 then maximum

The second derivative test is a method used to determine the concavity and extrema of a function. The second derivative of a function represents the rate of change of the function’s slope. If the second derivative is positive, the function is concave up, meaning it is curving upwards. If the second derivative is negative, the function is concave down, meaning it is curving downwards.

To perform the second derivative test, we follow these steps:
1. Find the critical points of the function by setting the first derivative equal to zero and solving for x.
2. Evaluate the second derivative at each critical point.
3. If the second derivative is positive at a critical point, the function has a relative minimum at that point.
4. If the second derivative is negative at a critical point, the function has a relative maximum at that point.
5. If the second derivative is zero, the test is inconclusive and another method must be used to determine the nature of the critical point.

For example, consider the function f(x) = x^3 – 3x^2 + 2x. We can find the first derivative by differentiating the function: f'(x) = 3x^2 – 6x + 2. To find the critical points, we set f'(x) equal to zero:

3x^2 – 6x + 2 = 0

Using the quadratic formula, we find that the solutions are x = 1 ± sqrt(1/3). Therefore, the critical points are (1 + sqrt(1/3), f(1 + sqrt(1/3))) and (1 – sqrt(1/3), f(1 – sqrt(1/3))).

Next, we find the second derivative by differentiating f'(x):

f”(x) = 6x – 6

Evaluating the second derivative at each critical point, we get:

f”(1 + sqrt(1/3)) = 2(sqrt(3)) > 0 (relative minimum)
f”(1 – sqrt(1/3)) = -2(sqrt(3)) < 0 (relative maximum) Therefore, f(x) has a relative minimum at x = 1 + sqrt(1/3) and a relative maximum at x = 1 - sqrt(1/3).

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