91Ó°ÊÓ

When studying relative extrema, the first step is to find the first derivative of the given function. The first derivative helps determine the rate of change of the function. For our function, which is $$G(x) = x^3 - 6x^2 + 10$$, we apply the power rule for differentiation.
The result is: \ G'(x) = 3x^2 - 12x

The first derivative, represented by \(G'(x)\), tells us how the function's slope changes over time. By setting \(G'(x)\) to zero, we find points called critical points where the slope is horizontal. To create a distinction, let's move on to understanding exactly what these critical points mean.

Critical points are where the first derivative of the function is zero or undefined. These points are important because they can indicate where relative maxima, minima, or inflection points occur. For \(G(x) = x^3 - 6x^2 + 10\), we find the critical points by solving \(G'(x) = 3x^2 - 12x = 0\).
Factoring this, we get: \3x(x - 4) = 0
The resulting critical points are \(x = 0\) and \(x = 4\). These values divide our function into separate intervals to further analyze.
But finding these points alone isn't enough; we need to understand the nature of these points using the second derivative.

The second derivative of a function helps determine the concavity of the graph at the critical points found earlier. For our function, we differentiate the first derivative \(G'(x)\) to get the second derivative:
\[G''(x) = 6x - 12\]
The second derivative test involves substituting the critical points into this equation. By doing so, we gain insights into whether these points are maxima, minima, or points of inflection.
Next, let's delve deeper into concavity and see what it tells us about our function at these critical points.

Concavity is a way to describe the curvature of a function. If the second derivative at a critical point is positive, the function is concave up (like a bowl) at that point, suggesting a local minimum. If the second derivative is negative, the function is concave down (like an upside-down bowl), indicating a local maximum.
For our critical points:

• At \(x = 0\),
  \(G''(0) = 6(0) - 12 = -12\), which is negative, suggesting a concave down shape and thus a local maximum.
• At \(x = 4\), \(G''(4) = 6(4) - 12 = 12\), which is positive, indicating a concave up shape and thus a local minimum.

Understanding the concavity helps us confirm the nature of our critical points. With this, let's identify the exact values of these relative maxima and minima.

A local maximum is a point where the function reaches a peak locally within an interval. At this point, the function has a higher value than any other point nearby. For our function, we identified that \(x = 0\) is a critical point with concave down shape, indicating a local maximum.

Now, we need to find the corresponding \(y\)-value:
\[G(0) = 0^3 - 6(0)^2 + 10 = 10\]
So, the local maximum occurs at the point \ (0, 10) \. This confirms that the function reaches a relative peak at this point.
Next, we will move to identifying the other type of extremum, the local minimum.

A local minimum is a point where the function attains a value that is lower than any other point nearby, essentially forming a valley. For our function, \(x = 4\) is a critical point indicating a concave up shape, which means a local minimum.

To find the corresponding \(y\)-value:
\[G(4) = 4^3 - 6(4)^2 + 10 = -38\]
Hence, the local minimum occurs at the point \(4, -38\). This gives us the lowest point in the function within the identified interval.

We have now identified the significant features of our function's relative extrema: the local maximum at \(0, 10)\) and the local minimum at \(4, -38\). With these key points, we can sketch a good representation of the function, showing its overall behavior.