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. \(f(x)=3 x^{5}-5 x^{3}\)

Step-by-Step Solution

Answer

Critical points: \(x = 0, 1, -1\); local min at \(x = 1\), local max at \(x = -1\). Inflection points at \(x = 0, \pm \frac{1}{\sqrt{2}}\).

Step 1: Find the First Derivative

Firstly, we need to differentiate the given function to find its first derivative. The function is given as:\[ f(x) = 3x^5 - 5x^3 \]Differentiating with respect to \(x\), we have:\[ f'(x) = \frac{d}{dx}(3x^5) - \frac{d}{dx}(5x^3) = 15x^4 - 15x^2 \]

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. Set the first derivative equal to zero and solve for \(x\):\[ 15x^4 - 15x^2 = 0 \]Factor out the common terms:\[ 15x^2(x^2 - 1) = 0 \]This results in:\[ x^2(15)(x^2 - 1) = 0 \]Thus, \(x = 0\) or \(x^2 - 1 = 0\). Solving the second equation gives \(x = \pm 1\). Hence, the critical points are \(x = 0, 1, -1\).

Step 3: Find the Second Derivative

Now, we need to find the second derivative to determine if there are any inflection points. Differentiate \(f'(x) = 15x^4 - 15x^2\) with respect to \(x\):\[ f''(x) = \frac{d}{dx}(15x^4) - \frac{d}{dx}(15x^2) = 60x^3 - 30x \]

Step 4: Find Inflection Points

Inflection points occur where the second derivative is zero or changes sign. Set the second derivative equal to zero:\[ 60x^3 - 30x = 0 \]Factor out the common terms:\[ 30x(2x^2 - 1) = 0 \]Thus, \(x = 0\) or \(2x^2 - 1 = 0\). Solving the second equation for \(x\) gives \(x = \pm\frac{1}{\sqrt{2}}\). Therefore, potential inflection points are \(x = 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\).

Step 5: Classify Critical Points Using Second Derivative Test

Use the second derivative to decide if each critical point is a local maximum, minimum, or neither:- For \(x = 0\), \(f''(0) = 0\), hence the second derivative test is inconclusive.- For \(x = 1\), \(f''(1) = 30\), which is positive, indicating a local minimum.- For \(x = -1\), \(f''(-1) = -30\), which is negative, indicating a local maximum.

Key Concepts

First Derivative

When we talk about the first derivative in calculus, we are referring to the rate at which a function is changing at any given point. This is also referred to as the slope of the tangent line to the curve of the function. For the function \[ f(x) = 3x^5 - 5x^3, \]calculating its first derivative means applying the power rule of differentiation. This rule states that for any function of the form \( ax^n \), the derivative is \( nax^{n-1} \). Applying this to each term of our function,

• The derivative of \( 3x^5 \) is \( 15x^4 \).
• The derivative of \( -5x^3 \) is \( -15x^2 \).

Combine these derivatives to find the first derivative of the original function:\[ f'(x) = 15x^4 - 15x^2. \]The next step is to find the critical points. Critical points occur where this first derivative is equal to zero or undefined. In this case, the derivative is defined everywhere, so we solve\[ 15x^4 - 15x^2 = 0 \]to find where changes in the curve's slope occur. This leads to the solutions \( x = 0, 1, -1 \) as your critical points. Simply put, these are values where the slope of the function becomes zero, indicating a peak, trough, or a point of inflection.

Second Derivative

The second derivative provides us with more insights about the curvature of the function. Calculating it involves differentiating the first derivative. The second derivative, \[ f''(x) = 60x^3 - 30x, \]determines the concavity of the function. This means it tells us whether the curve is concave up (shaped like a cup) or concave down (shaped like a cap) at a given point.

To find inflection points, we look at where the second derivative is zero or changes sign. Solving \[ 60x^3 - 30x = 0, \]we find possible inflection points at \( x = 0, \pm \frac{1}{\sqrt{2}} \). These are the points where the function might change its concavity direction.

The second derivative also supports the classification of critical points found using the first derivative. By determining the sign of the second derivative at these critical points:

• if \( f''(x) > 0 \), the function is concave up, suggesting a local minimum.
• if \( f''(x) < 0 \), the function is concave down, suggesting a local maximum.
• if \( f''(x) = 0 \), it might be an inflection point, or further investigation is needed.

For our example, at \( x = 1 \), the second derivative is positive, indicating a local minimum. At \( x = -1 \), it is negative, suggesting a local maximum.

Inflection Points

Inflection points are fascinating places where the function changes its concavity. In simpler terms, the curve goes from being "bowl-shaped" to "cap-shaped" or vice versa. To uncover inflection points, we work with the second derivative. When solving \[ 60x^3 - 30x = 0, \]we factor it as \[ 30x(2x^2 - 1) = 0, \]leading us to potential inflection points at \( x = 0, \pm \frac{1}{\sqrt{2}} \).

However, merely finding x-values where the second derivative is zero is not enough. The real inflection point must be where the sign of the second derivative actually changes. That means you have to check nearby values of x to ensure the concavity direction shifts.

Let’s consider our example to illustrate:

• At \( x = 0 \), if we check just before and after, the second derivative remains positive, indicating no change in concavity.
• Conversely, at \( x = \pm \frac{1}{\sqrt{2}} \), if the concavity shifts from positive to negative or vice versa, these are true inflection points.

This exploration helps in graphing the function accurately, showing the nuanced transitions in curvature and thus depicting the true shape of the graph.