91Ó°ÊÓ

A derivative measures how a function changes as its input changes. Think of it as the function's sensitivity. In simpler terms, it's like asking, "If I wiggle the input slightly, how will the output wiggle in response?" In our optimization problem, the function to consider is

• \( f(x) = x - x^2 \)

Here, the derivative, denoted as \( f'(x) \), is the mathematical tool we need to understand the rate of change. We find it by differentiating:

• \( f'(x) = 1 - 2x \)

This formula tells us how "steep" or "flat" the curve of the function is at any point \( x \).
When \( f'(x) = 0 \), the rate of change is zero indicating a potential maximum or minimum point. This is why derivatives are crucial for finding the peaks and valleys in functions.

Critical points are where the derivative of a function equals zero or where it's undefined. They are like signposts telling us to "stop and look around" because something interesting is happening. For our function, \( f(x) = x - x^2 \), we set the derivative equal to zero:

• \( 1 - 2x = 0 \)
• Solve for \( x \), and we find \( x = \frac{1}{2} \)

This \( x \) value is a critical point. Don't skip verifying if it actually corresponds to a maximum or minimum without further testing.
It's like finding a fork in the road; you need to determine which way leads to the top of the hill.

The Second Derivative Test helps us determine whether a critical point is a local maximum, a local minimum, or a point of inflection. In this context, it acts like a detective tool to uncover the nature of the critical point found.
For our function, the second derivative is:

• \( f''(x) = -2 \)

With the second derivative test, if \( f''(x) < 0 \), the graph is concave down at that point, indicating a local maximum.
In our problem, because \( f''(x) = -2 \) is negative, we confirm that the function \( x - x^2 \) reaches a maximum at \( x = \frac{1}{2} \).
This test gives us the confidence that we've pinpointed the highest spot on our "function hill."