Problem 22 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 21 through 24 , find all critical numbers for the given function \(f(x)\) and use the second derivative test to determine which (if any) critical points are relative maxima or relative minima. $$ f(x)=x(2 x-3)^{2} $$

Step-by-Step Solution

Answer

Critical points are at \(x = \frac{3}{4}\ and \(x = \frac{3}{2}. \(x = \frac{3}{2}\) is a relative minimum. Second derivative test is inconclusive for \(x = \frac{3}{4}.\)

Step 1: Find the first derivative

To find the critical points, take the first derivative of the function. Given the function \(f(x) = x(2x-3)^2,\) use the product rule: \[f'(x) = (2x - 3)^2 + x \times 2(2x - 3).\] Simplify it to get: \[f'(x) = (2x - 3)^2 + 2x(2x - 3).\] This will further simplify to: \[f'(x) = (2x - 3)[2x - 3 + 2x] = (2x - 3)(4x - 3).\]

Step 2: Find critical points

Set the first derivative equal to zero to find critical points: \(f'(x) = (2x-3)(4x-3) = 0.\) Therefore, \(2x - 3 = 0\) or \(4x - 3 = 0.\) Solving these equations, we get \[x = \frac{3}{2} \text{ and } x = \frac{3}{4}.\]

Step 3: Find the second derivative

Take the second derivative of the original function: The simplified first derivative was \(f'(x) = (2x-3)(4x-3).\) Use the product rule again: \[f''(x) = (2x-3)'(4x-3) + (2x-3)(4x-3)'.\] Simplify to get: \[f''(x) = 2(4x-3) + (2x-3)4 = 8x - 6 + 8x - 6 = 16x - 12.\]

Step 4: Apply the second derivative test

Substitute the critical points into the second derivative to classify them: \[f''\left(\frac{3}{4}\right) = 16\left(\frac{3}{4}\right) - 12 = 12 - 12 = 0.\] Thus, the second derivative test is inconclusive at \(x = \frac{3}{4}.\)

Step 5: Evaluate the second critical point

Substitute the other critical point: \[f''\left(\frac{3}{2}\right) = 16\left(\frac{3}{2}\right) - 12 = 24 - 12 = 12 > 0.\] Since \(f''\left(\frac{3}{2}\right) > 0,\) the function is concave up at \(x = \frac{3}{2},\) indicating a relative minimum.

Key Concepts

Understanding Critical Points

Critical points are places on the graph of a function where the first derivative is zero or undefined. These points are important because they may indicate local maxima, minima, or points of inflection. To find them, we set the first derivative equal to zero and solve for the variable. In the given function, we identified critical points by solving \((2x - 3)(4x - 3) = 0\), which yielded \(-x = \frac{3}{2} \) and \(x = \frac{3}{4}.\)

First Derivative for Critical Points

To find critical points, we must first take the derivative of the function. The first derivative tells us the slope of the function at any given point. In the solution, we used the product rule to differentiate the original function \(f(x) = x(2x-3)^2.\) The product rule states: \( (uv)' = u'v + uv'.\)
Applying this to our function, we derived \(f'(x)\) to be \( (2x-3)^2 + x \cdot 2(2x-3)\), simplifying further to \((2x - 3)(4x - 3).\) We then set this equal to zero to find the critical points.

Relative Extrema

Relative extrema refer to local minima and maxima of a function. To classify critical points as relative extrema, we can use the second derivative test. If the second derivative at a critical point is positive, the function has a relative minimum there. If it's negative, there is a relative maximum. Otherwise, the test is inconclusive. In our problem, at \(x = \frac{3}{2}\), the function's second derivative was greater than zero, indicating a relative minimum.

Using the Product Rule

The product rule is used when differentiating a product of two functions. For a function \(f(x) = u(x) \cdot v(x),\) the product rule is written as: \(f'(x) = u'(x)v(x) + u(x)v'(x).\) This rule was applied in taking the derivative of \(f(x) = x(2x-3)^2.\) By letting \(u(x) = x\) and \(v(x) = (2x-3)^2,\) we calculated \(u'(x) = 1\) and \(v'(x) = 2(2x-3) \cdot 2 = 4(2x-3).\)

Examining Concavity and Second Derivative Test

Concavity refers to whether a function curves upwards or downwards. The second derivative of a function gives information on its concavity. If the second derivative, \(f''(x),\) is positive at a point, the function is concave up at that point; if negative, concave down. We use this to apply the second derivative test. For \(f''(x) = 16x - 12,\) substituting \(x = \frac{3}{2},\) we found \(f''(x) = 12,\) illustrating concave up and confirming a relative minimum.