The First Derivative Test Made Simple

The First Derivative Test is a helpful tool in calculus. It helps us find where a function reaches high or low points, called local extrema. To use this test, we need to understand some basic ideas about derivatives and critical points.

What Are Critical Points?

Let's start with critical points. A function ( f(x) ) has a critical point at ( x = c ) if:

1. The derivative ( f'(c) = 0 ), or
2. The derivative ( f'(c) ) cannot be calculated (it is undefined).

Critical points are important because they tell us where the function might reach a local maximum (highest point) or a local minimum (lowest point).

To find critical points, we first need to calculate the first derivative of the function and see where it equals zero or is undefined.

For example, let’s look at the function ( f(x) = x^3 - 3x^2 + 4 ).

First, we find its derivative:

$f'(x) = 3x^2 - 6.$

Next, we set the derivative equal to zero:

$3x^2 - 6 = 0 \implies x^2 = 2 \implies x = \pm\sqrt{2}.$

So, our critical points are ( x = \sqrt{2} ) and ( x = -\sqrt{2} ).

The First Derivative Test Explained

The First Derivative Test helps us figure out if these critical points are local maximums, local minimums, or neither.

Here’s how we use the First Derivative Test:

1. Find the critical points: These are where ( f'(x) = 0 ) or is undefined.

2. Make a number line: Draw a number line and mark the critical points ( -\sqrt{2} ) and ( \sqrt{2} ) on it.

3. Choose test points: Pick points to test in the sections created by the critical points. You should choose a point on the left and right of each critical point.

4. Check the sign of the derivative:

   • If ( f'(x) ) changes from positive to negative at a critical point ( c ), then ( f(c) ) is a local maximum.
   • If ( f'(x) ) changes from negative to positive at ( c ), then ( f(c) ) is a local minimum.
   • If ( f'(x) ) doesn’t change, then ( c ) is neither a maximum nor minimum.

Now, let’s test our example using ( f'(x) = 3x^2 - 6 ) and the critical points ( -\sqrt{2} ) and ( \sqrt{2} ).

• Testing intervals:

  • For the interval ( (-\infty, -\sqrt{2}) ), let’s use ( x = -3 ).
  • For the interval ( (-\sqrt{2}, \sqrt{2}) ), let’s use ( x = 0 ).
  • For the interval ( (\sqrt{2}, \infty) ), let’s use ( x = 3 ).

• Evaluate ( f'(x) ):

  • For ( x = -3 ): $f'(-3) = 3(-3)^2 - 6 = 27 - 6 = 21 > 0$ The derivative is positive before ( -\sqrt{2} ).

  • For ( x = 0 ): $f'(0) = 3(0)^2 - 6 = -6 < 0$ The derivative is negative between ( -\sqrt{2} ) and ( \sqrt{2} ).

  • For ( x = 3 ): $f'(3) = 3(3)^2 - 6 = 27 - 6 = 21 > 0$ The derivative is positive after ( \sqrt{2} ).

What Do the Results Mean?

Now, let’s summarize what we learned about the critical points:

• At ( x = -\sqrt{2}), the derivative changed from positive to negative. This means ( f(-\sqrt{2}) ) is a local maximum.

• At ( x = \sqrt{2}), the derivative changed from negative to positive. This means ( f(\sqrt{2}) ) is a local minimum.

In simple terms, the First Derivative Test is a clear way to find local extrema in calculus. By identifying critical points and looking at how the first derivative behaves around them, we can understand how a function acts in different parts.

Why Is This Important?

Knowing about local extrema using the First Derivative Test matters in many areas, including business, science, and engineering. People often want to find the highest profit, lowest cost, or best solutions.

So, the First Derivative Test is not just a theory; it’s a practical tool. By using this test, we can learn more about how functions behave, paving the way for deeper studies in calculus and its many uses in real life.