Problem 20 - Applied Calculus

Question

Use the first derivative to find all critical points and use the second derivative to find all inflection points. Use a graph to identify each critical point as a local maximum, a local minimum, or neither. $$f(x)=x^{5}-5 x^{4}+35$$

Step-by-Step Solution

Answer

Critical points: \( x = 0\) (min), \( x = 4\) (max); Inflection point: \( x = 3 \).

Step 1: Find the First Derivative

To find the critical points of the function, we first need to determine its first derivative. Given the function \( f(x) = x^{5} - 5x^{4} + 35 \), let's differentiate it with respect to \( x \):\[ f'(x) = 5x^{4} - 20x^{3} \]

Step 2: Solve for Critical Points

Critical points occur where the first derivative is zero or undefined. Set the first derivative \( f'(x) = 5x^{4} - 20x^{3} \) to zero and solve for \( x \):\[ 5x^{4} - 20x^{3} = 0 \]\[ 5x^{3}(x - 4) = 0 \]Factor out the common term:\[ x^{3}(x - 4) = 0 \]This gives us the solutions \( x = 0 \) and \( x = 4 \). These are the critical points.

Step 3: Find the Second Derivative

To find potential inflection points, we calculate the second derivative. Differentiate the first derivative \( f'(x) = 5x^{4} - 20x^{3} \) with respect to \( x \):\[ f''(x) = 20x^{3} - 60x^{2} \]This is the second derivative.

Step 4: Solve for Inflection Points

Inflection points occur where the second derivative is zero or undefined. Set \( f''(x) = 20x^{3} - 60x^{2} \) to zero and solve for \( x \):\[ 20x^{2}(x - 3) = 0 \]This simplifies to \( x = 0 \) and \( x = 3 \) as solutions. These are the potential inflection points.

Step 5: Use a Graph to Classify Critical Points

To determine whether the critical points are local maxima, minima, or neither, we can plot the function and its derivatives. Observing the graph, we analyze the slope changes:- At \( x = 0 \), the derivative changes sign from negative to positive, indicating a local minimum.- At \( x = 4 \), the derivative changes sign from positive to negative, indicating a local maximum.

Key Concepts

First Derivative

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.

Second Derivative

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.

Local Maxima and Minima

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

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.