The first derivative test is used in calculus to determine whether a critical point is a local minimum, local maximum, or neither. This is done in three steps:

**Step 1 – Identify the Critical Points**

The way we identify the critical points for the first derivative test is to take the derivative of a function, set the function equal to zero, and find the points that satisfy the function. For example, lets say our function is f(x) = (1/3)x^{3}-2x^{2}+3x-2. First we take the derivative, giving us f’(x) = x^{2}-4x+3. Next we set the function equal to zero as 0 = 6x^{2}-6x+4, then simplifying as 0 = (x-3)(x-1). This makes our critical points x=1 and x=3.

**Step 2: Determine behaviour around the critical points**

The way we determine behaviour around the critical points is by plugging values below and above the critical points into the first derivative function. If the first derivative function is positive then the slope is positive and f(x) is increasing. When the derivative function is negative then slope is negative and f(x) is decreasing. To use the example above, we have two critical points, x=1 and x=3. So, we can plug the values x=0, 2, and 4 to determine the behaviour of the function. In this case, when x=0 then f’(x) is positive, when x=2 then f’(x) is negative, and when x=4 then f’(x) is positive. This means that slope is positive until it gets to x=1, then negative until it gets to x=3, then is positive at all values after that.

**Step 3: Evaluate the Critical Points**

Now that we know the behaviour around the critical points, we’re able to determine whether each point is a local maximum, local minimum, or neither. Since we know that a local maximum has a positive slope to the left and a negative slope to the right, we know that the point x=1 is a local maximum. Also, since we know that a local minimum has a negative slope to the left and a positive slope to the right, we know that x=3 is a local minimum. While we didn’t have a critical point that was neither a local maximum nor a local minimum, we would have known that we had one if the critical point had either a positive slope on both sides or a negative slope on both sides.

By following these three steps, you shouldn’t have any difficulty solving optimization problems, or any other problems requiring the first derivative test.

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