91Ó°ÊÓ

In calculus, a derivative represents the rate at which a function changes at a given point. It's like measuring how fast or slow something is moving along a curve. For the function \( f(x) = 3 - x - \frac{1}{2}x^2 \), the derivative \( f'(x) \) is found by applying basic differentiation rules. This involves differentiating each term individually. The derivative of a constant (like 3) is zero. The derivative of \( -x \) is \(-1\), and the derivative of \(-\frac{1}{2}x^2\) is \(-x\). Together, this gives us:

• \( f'(x) = -1 - x \)

This derivative \( f'(x) \) helps us understand how the original function \( f(x) \) is changing. Once we find this, we can move towards calculating the critical points of the function.

Critical points of a function occur where its derivative equals zero or is undefined. These points are like hints showing where the function might have a peak or valley. We find the critical points by setting the derivative \( f'(x) = -1 - x \) to zero and solving for \( x \). This yields:

• \( -1 - x = 0 \)
• Solve for \( x \) to get \( x = -1 \)

When we solve the equation, \( x = -1 \) becomes our critical point. It's important because it might lead us to discover maximum or minimum values of the function. Next, we need to find out if it's a point of a local maximum, minimum, or something else.

The Second Derivative Test is a method to determine whether a critical point is a maximum, minimum, or a saddle point. Once the critical point is found, we calculate the second derivative \( f''(x) \) of the function. For our function:

• \( f''(x) = -1 \)

Since \( f''(x) = -1 \) is less than zero, this indicates that the function is concave down at \( x = -1 \). Thus, the critical point \( x = -1 \) is actually a point of local maximum. This is because concave down functions tend to have peaks rather than troughs at critical points.

Finding maximum and minimum values of a function helps us understand the highest and lowest points that a curve can reach. Once we've identified a critical point as a maximum, we can find the actual maximum value by plugging that point back into the original function \( f(x) \). For the function \( f(x) = 3 - x - \frac{1}{2}x^2 \), substituting \( x = -1 \) gives us:

• \( f(-1) = 3 - (-1) - \frac{1}{2}(-1)^2 \)
• Calculate this to get \( f(-1) = 3.5 \)

The maximum value of the function is thus 3.5 when \( x = -1 \). This concludes our journey through the calculus optimization of \( f(x) \), where identifying derivatives and applying tests help uncover critical features of the function.