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A local maximum of a function is a point where the function value is greater than the values at points nearby. Essentially, it's a peak in a specific area of a graph. When you're standing at a local maximum, the function elevates to this point and then starts declining immediately in the neighborhood. When attempting to find a local maximum, the first step is usually to find critical points by setting the first derivative to zero. A critical point can become a local maximum if the function's derivative changes from positive to negative at that point. This signifies the function was increasing and then began decreasing, hence reaching a peak at that critical point.In the example you are studying, the function has a local maximum at \(x = \frac{3}{7}\), as verified by examining the sign of the first derivative around this point. The derivative transitions from positive before \(x = \frac{3}{7}\) to negative afterwards.

A local minimum is the opposite of a local maximum. It's a dip or a trough in the graph of a function, where the function value is less than all nearby values. At this specific point, the function appears to sink and then rise again as you move away.To determine whether a critical point is a local minimum, you look for where the first derivative changes sign from negative to positive. This means the function was dipping down and then begins rising, indicating a local low or trough.For the function given in your problem, there are local minima at \(x = 0\) and \(x = 1\). By evaluating changes in the first derivative around these points, we confirm that the derivative transitions from negative to positive, assuring us of their classification as local minima.

The first derivative of a function, represented as \(f'(x)\), is crucial in finding the slope or rate of change of the function at any point \(x\). It's instrumental in identifying critical points, which are potential candidates for local extrema (maxima or minima).
In your exercise, the first derivative is found using the product rule. The function \(f(x) = x^3 (1-x)^4\) results in a first derivative of \(f'(x) = x^2 (1-x)^3 (3 - 7x)\) after simplification. Critical points are determined by setting this derivative equal to zero, as they represent locations where the function's slope is either zero or undefined, possibly indicating peaks or troughs.
Understanding the first derivative helps assess where the function increases or decreases, paving the path toward classifying each critical point correctly.

The second derivative, denoted as \(f''(x)\), provides insights into the concavity of a function. In context, it tells us whether the graph of a function curves "upward" or "downward" at a specific point.
While looking for local extrema, the second derivative test can be used. If \(f''(x) > 0\) at a critical point, the function is concave up, and the point is a local minimum. Conversely, if \(f''(x) < 0\), the function is concave down, and it indicates a local maximum.
Although directly calculating the second derivative may be complex, especially for involved functions like \( f(x) = x^3 (1-x)^4 \), evaluating the sign changes around critical points provides a good alternative method to confirm the nature of maxima and minima.

The product rule in calculus is a method for differentiating products of two functions. Written as \((uv)' = u'v + uv'\), this rule helps find the derivative of functions expressed as products of simpler functions.
In the exercise at hand, \(f(x) = x^3(1-x)^4\) required the use of the product rule. By defining \(u = x^3\) and \(v = (1-x)^4\), their derivatives, \(u' = 3x^2\) and \(v' = -4(1-x)^3\), were calculated individually. The product rule was then applied, combining these derivatives to obtain the first derivative: \(f'(x) = 3x^2(1-x)^4 - 4x^3(1-x)^3\).
Grasping the product rule is essential for handling functions where different components multiply together, providing necessary insights into how they jointly influence the derivative.