Problem 234 - Calculus Volume 1

Question

Determine a. intervals where \(f\) is increasing or decreasing, b. local minima and maxima of \(f\), c. intervals where \(f\) is concave up and concave down, and d. the inflection points of \(f\). Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator. ] \(f(x)=(x-2)^{2}(x-4)^{2}\)

Step-by-Step Solution

Answer

Intervals of increase: \((3, \, 4)\) and decrease: \((2, \, 3)\). Local min at \(x=4\), max at \(x=2\). Concave up: \(([3, \, 4])\), concave down: \((\text{others})\). Inflection at \(x=3\).

Step 1: Find Critical Points

To find the critical points of the function \(f(x)=(x-2)^{2}(x-4)^{2}\), we first find its derivative. Apply the product rule: Since \(f(x) = (x-2)^2 \cdot (x-4)^2\), we differentiate to get \(f'(x) = 2(x-2)(x-4)^2 + 2(x-4)(x-2)^2\). Simplify this to find the values of \(x\) where \(f'(x) = 0\).

Step 2: Solve for Critical Points

Set \(f'(x) = 0\) and solve for \(x\). Simplifying gives \(f'(x) = 2(x-2)(x-4)[(x-4) + (x-2)] = 2(x-2)(x-4)(2x-6) = 0\). Solve each factor: \(x-2 = 0\) or \(x-4 = 0\) or \(2x-6=0\). Thus, the critical points are \(x=2, x=4, x=3\).

Step 3: Determine Increasing/Decreasing Intervals

Use the critical points to find intervals: \((-\infty, 2)\), \((2, 3)\), \((3, 4)\), and \((4, \infty)\). Test a point in each interval to determine the sign of \(f'(x)\) to see if the function is increasing or decreasing in that interval.

Step 4: Determine Local Minima and Maxima

Based on the sign changes of \(f'(x)\) around the critical points, determine if they are local minima or maxima. Check the behavior of \(f(x)\) as \(x\) moves through these critical points. If \(f'(x)\) changes from positive to negative, there is a local maximum. If it changes from negative to positive, there is a local minimum.

Step 5: Find Concavity Intervals

Compute the second derivative \(f''(x)\) to determine the concavity. Simplifying \(f''(x)\) and solving \(f''(x) = 0\) will give potential inflection points. Use \(f''(x)\) to test intervals to see where \(f(x)\) is concave up (\(f''(x) > 0\)) or concave down (\(f''(x) < 0\)).

Step 6: Determine Inflection Points

Inflection points occur where \(f''(x)\) changes signs. By analyzing the intervals of concavity and confirming sign changes in \(f''(x)\), identify any such points.

Key Concepts

Critical Points

Critical points are vital in finding key characteristics of a function. They occur at values where the function's derivative is zero or undefined. For the function \( f(x)=(x-2)^{2}(x-4)^{2} \), finding the derivative follows the product rule. This gives us \( f'(x) = 2(x-2)(x-4)^2 + 2(x-4)(x-2)^2 \). Simplifying further, we find \( f'(x) = 2(x-2)(x-4)(2x-6) \). The critical points are where this derivative equals zero: \( x = 2, 3, 4 \). These are the points at which the curve can potentially change direction from increasing to decreasing or vice versa. Analyzing these allows us to understand the behavior of the graph.

Derivative Analysis

The analysis of the derivative \( f'(x) \) helps determine which intervals the function is increasing or decreasing. With the critical points found as \( x=2, x=3, \) and \( x=4 \), we divide the number line into intervals:

• \((-\infty, 2)\)
• \((2, 3)\)
• \((3, 4)\)
• \((4, \infty)\)

For each interval, we pick a test point such as \( x = 1, 2.5, 3.5, \) and \( 5 \) respectively to evaluate the sign of \( f'(x) \). This will tell us if the function is increasing (\( f'(x) > 0 \)) or decreasing (\( f'(x) < 0 \)) within each section. This critical evaluation paves the way for understanding where the function exhibits local extrema.

Concavity and Inflection Points

Understanding concavity is crucial to revealing more about a function's graph. It reflects how the slope of a curve changes. For \( f(x)=(x-2)^2(x-4)^2 \), we examine the second derivative, \( f''(x) \). Calculating \( f''(x) \) involves differentiating \( f'(x) \), which offers insights on where the graph is concave up (bowed upwards) or concave down (bowed downwards). Specifically:

• \( f''(x) > 0 \) indicates concave up.
• \( f''(x) < 0 \) indicates concave down.

Inflection points occur at \( x \) values where \( f''(x) \) changes signs. These points signal where the curvature changes direction, which is visually represented as a change from convex to concave or vice versa.

Local Minima and Maxima

Finding local minima and maxima is a fundamental application of critical points. These are points where the function achieves local peaks or troughs. After determining the critical points and evaluating \( f'(x) \) around these points, we classify them:

• A local maximum occurs if \( f'(x) \) changes from positive to negative.
• A local minimum occurs if \( f'(x) \) changes from negative to positive.

In the example \( f(x)=(x-2)^2(x-4)^2 \), this step helps identify where the graph has crests and troughs, informing optimal points within specified intervals. By understanding these transitions, we gain a more accurate visualization of the function's behavior over its domain.