91Ó°ÊÓ

The first derivative of a function is essential for finding critical points. These points are where the function potentially changes direction, i.e., from increasing to decreasing or vice-versa. A critical point can occur when the first derivative is zero or undefined.
In the exercise provided, the function is given by

• \(f(x) = x^{5} - 5x^{4} + 35\)

Differentiating this with respect to \(x\) yields the first derivative:

• \(f'(x) = 5x^{4} - 20x^{3}\)

To find the critical points, we set \(f'(x)\) equals zero: \[5x^{4} - 20x^{3} = 0\] Solving this equation, we factor out common terms to obtain: \[x^{3}(x - 4) = 0\] Therefore, the solutions are \(x = 0\) and \(x = 4\). These values of \(x\) are the critical points of the function.

The second derivative informs us about the concavity of the function; in other words, it helps us find potential inflection points. Inflection points are where the function changes concavity, from concave up to concave down, or vice-versa.
In our case, we begin by taking the derivative of \(f'(x) = 5x^{4} - 20x^{3}\). Differentiating this, we obtain the second derivative:

• \(f''(x) = 20x^{3} - 60x^{2}\)

To locate inflection points, we set \(f''(x)\) to zero and solve:

• \(20x^{2}(x - 3) = 0\)

The solutions to this equation are \(x = 0\) and \(x = 3\). These are the potential inflection points, signifying where the curve could change its concavity.

Determining whether a critical point is a local maximum or minimum is fundamental in understanding the behavior of a function. By analyzing the sign of the first derivative around these critical points, one can determine their nature.
In the context of our exercise, we already found critical points at \(x = 0\) and \(x = 4\). We examine the behavior of \(f'(x)\) around these points:

• At \(x = 0\), the derivative changes from negative to positive. This change indicates a local minimum.
• At \(x = 4\), the derivative changes from positive to negative, marking a local maximum.

Such analyses help us visualize and understand how the function behaves near these particular \(x\) values.

Inflection points provide insights into how the function’s curvature changes. The second derivative is a valuable tool to detect these points since they reveal where the function’s concavity shifts.
For the function given, potential inflection points occur where the second derivative:\[f''(x) = 20x^{3} - 60x^{2} = 0\]This results in \(x = 0\) and \(x = 3\), suggesting these might be inflection points.
Remember, to classify these as true inflection points, the second derivative must change signs around these \(x\) values. In practice, we may need to plot the graph or test the second derivative on either side of these points to confirm a sign change, further affirming shifts in the function's concavity. This understanding aids in fully picturing the function's shape and behavior.