Problem 14 - 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. $$f(x)=2 x^{3}+3 x^{2}-36 x+5$$

Step-by-Step Solution

Answer

Critical points: local maximum at \( x = -3 \), local minimum at \( x = 2 \). Inflection point at \( x = -\frac{1}{2} \).

Step 1: Find the First Derivative

To find the critical points, start by computing the first derivative of the given function, \( f(x) = 2x^3 + 3x^2 - 36x + 5 \). Use the power rule for differentiation: \( \frac{d}{dx}[x^n] = nx^{n-1} \). The first derivative is \( f'(x) = 6x^2 + 6x - 36 \).

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find the critical points: \( 6x^2 + 6x - 36 = 0 \). Simplify by dividing the entire equation by 6: \( x^2 + x - 6 = 0 \). Factor the quadratic equation into \( (x + 3)(x - 2) = 0 \). Thus, the critical points are \( x = -3 \) and \( x = 2 \).

Step 3: Classify Critical Points Using the Second Derivative

Compute the second derivative of the function, \( f''(x) = 12x + 6 \). Evaluate the second derivative at each critical point to classify them. \( f''(-3) = 12(-3) + 6 = -30 \), which is less than zero, indicating a local maximum at \( x = -3 \). \( f''(2) = 12(2) + 6 = 30 \), which is greater than zero, indicating a local minimum at \( x = 2 \).

Step 4: Find Inflection Points

Inflection points occur where the second derivative changes sign. Set \( f''(x) = 12x + 6 = 0 \). Solving for \( x \), we get \( x = -\frac{1}{2} \). Test the intervals around \( x = -\frac{1}{2} \) to ensure a sign change in \( f''(x) \) occurs, confirming an inflection point.

Step 5: Graph the Function

Graph \( f(x) = 2x^3 + 3x^2 - 36x + 5 \). Use the information from the first and second derivatives: the graph should show a local maximum at \( x = -3 \), a local minimum at \( x = 2 \), and an inflection point at \( x = -\frac{1}{2} \). This visual confirmation suggests that the calculations align with changes in direction and concavity on the graph.

Key Concepts

Critical Points

Critical points are special values of the variable where the behavior of a function changes significantly. These points can either be a local maximum, minimum, or sometimes, neither.
They occur where the first derivative of a function is zero or undefined.Let's dive into it with our function:

• Function given: \( f(x) = 2x^3 + 3x^2 - 36x + 5 \)
• The first derivative is computed as: \( f'(x) = 6x^2 + 6x - 36 \)

By setting \( f'(x) = 0 \), we solve for \( x \) in the equation: \( 6x^2 + 6x - 36 = 0 \)
This can be simplified to \( x^2 + x - 6 = 0 \), which factors into \( (x+3)(x-2) = 0 \). Therefore, the critical points are \( x = -3 \) and \( x = 2 \).
These critical points indicate potential peaks or valleys in the function graph.

First Derivative

The first derivative of a function is key in the study of calculus. It tells us the slope of the tangent line at any point on a curve. This is significant for determining where the function is increasing or decreasing.
For our function \( f(x) = 2x^3 + 3x^2 - 36x + 5 \):

• First derivative is \( f'(x) = 6x^2 + 6x - 36 \)

The zeros of this derivative give us the critical points, which can be maximum, minimum, or a saddle point.
By analyzing these points, we can understand the turning behavior of the graph around these points.

Second Derivative

The second derivative provides insight into the concavity of the function. While the first derivative gives slope information, the second derivative shows how that slope is changing.
For instance, in our function:

• Second derivative is \( f''(x) = 12x + 6 \)

We use this to determine the nature of our critical points derived from the first derivative.
Evaluating \( f''(x) \) at the critical points:

• \( f''(-3) = -30 \), which indicates a local maximum because it is negative.
• \( f''(2) = 30 \), suggesting a local minimum as it's positive.

Thus, the second derivative plays a vital role in classifying the critical points effectively.

Inflection Points

Inflection points are where the function changes concavity – from concave up to concave down, or vice versa. This means the second derivative changes sign at these points.
To find inflection points for the function:

• Set \( f''(x) = 12x + 6 = 0 \)

Solving gives \( x = -\frac{1}{2} \). To confirm, examine intervals around this \( x \) value to ensure there is a sign change in \( f''(x) \).
This confirmatory step is crucial as it verifies that the function indeed has an inflection point at \( x = -\frac{1}{2} \). This means the function's curve changes its upward or downward bending at this point.