Problem 72 - Calculus Early Transcendentals: Pearson New International Edition

Question

Draw the graphs of \(f(x)=x^{3}-4 x^{2}+3\) and its derivative \(f^{\prime}(x)\) on the interval \([-2,5]\) using the same axes. (a) Where on this interval is \(f^{\prime}(x)<0\) ? (b) Where on this interval is \(f(x)\) decreasing? (c) Make a conjecture. Experiment with other intervals and other functions to support this conjecture.

Step-by-Step Solution

Answer

(a) \((0, \frac{8}{3})\). (b) \((0, \frac{8}{3})\). (c) A function decreases where its derivative is negative.

Step 1: Find the derivative of the function

To find where the function is increasing or decreasing, we first need to find its derivative. The given function is \( f(x) = x^3 - 4x^2 + 3 \). To find its derivative, apply the power rule: \( f'(x) = 3x^2 - 8x \).

Step 2: Solve the derivative for critical points

To find critical points where the function might change from increasing to decreasing (or vice versa), set \( f'(x) = 0 \): \( 3x^2 - 8x = 0 \). Factoring the equation gives: \( x(3x - 8) = 0 \), resulting in critical points \( x = 0 \) and \( x = \frac{8}{3} \).

Step 3: Determine where \( f'(x) < 0 \)

Check the intervals determined by the critical points to see where \( f'(x) \) is negative. Test intervals: \((-\infty, 0)\), \((0,\frac{8}{3})\), and \((\frac{8}{3}, \infty)\): - For \( x < 0 \), choose \( x = -1 \): \( f'(-1) = 3(-1)^2 - 8(-1) = 3 + 8 = 11 > 0 \) (positive)- For \( 0 < x < \frac{8}{3} \), choose \( x = 1 \): \( f'(1) = 3(1)^2 - 8(1) = 3 - 8 = -5 < 0 \) (negative)- For \( x > \frac{8}{3} \), choose \( x = 3 \): \( f'(3) = 3(3)^2 - 8(3) = 27 - 24 = 3 > 0 \) (positive)Thus, \( f'(x) < 0 \) on the interval \((0, \frac{8}{3})\).

Step 4: Determine where \( f(x) \) is decreasing

A function \( f(x) \) is decreasing where its derivative is negative. Based on Step 3, \( f(x) \) is decreasing on the interval \((0, \frac{8}{3})\).

Step 5: Formulate a conjecture

Given that \( f(x) \) is decreasing on the interval \((0, \frac{8}{3})\) where \( f'(x) < 0 \), we can conjecture that a function decreases on intervals where its derivative is negative. This can be tested with other functions and intervals.

Key Concepts

Derivatives

Derivatives play a crucial role in calculus, allowing us to determine the rate at which a function changes. To find the derivative of a function like \( f(x) = x^3 - 4x^2 + 3 \), we use differentiation rules. The power rule is particularly useful here: for any term of the form \( ax^n \), the derivative is \( anx^{n-1} \). Applying this to our function gives \( f'(x) = 3x^2 - 8x \).

With derivatives, we can explore how a function behaves. They provide insights on functions' slopes at various points, telling us if a function is increasing, decreasing, or remaining constant. By examining \( f'(x) \), we can determine these characteristics over specified intervals.

In summary, derivatives are foundational for understanding how functions evolve, enabling us to predict and interpret different behaviors based on their mathematical models.

Critical Points

Critical points are where a function's derivative equals zero or becomes undefined. They're the locations where the function might have local maxima, minima, or saddle points.

For the function \( f(x) = x^3 - 4x^2 + 3 \), the derivative is \( f'(x) = 3x^2 - 8x \). To find critical points, solve \( f'(x) = 0 \):

• \( 3x^2 - 8x = 0 \)
• Factorizing gives \( x(3x - 8) = 0 \)
• The critical points are \( x = 0 \) and \( x = \frac{8}{3} \)

These points indicate where the function's slope transitions between increasing or decreasing. However, not all critical points correspond to extremes; some might be inflection points where the slope changes its direction.

Analyzing critical points helps in sketching the graph of a function and understanding its behavior more clearly, particularly in identifying segments of increase or decrease.

Increasing and Decreasing Functions

Understanding whether a function is increasing or decreasing within a certain interval is essential in calculus. A function is increasing if its derivative is positive across an interval, while it is decreasing if the derivative is negative.

Let's consider our function \( f(x) = x^3 - 4x^2 + 3 \). After finding its derivative \( f'(x) = 3x^2 - 8x \), we determine the sign of the derivative in various intervals.

• From Step 3 of the original solution, we found that \( f'(x) < 0 \) on the interval \( (0, \frac{8}{3}) \)
• Thus, \( f(x) \) is decreasing in this interval

By segmenting the intervals between critical points, one can effectively determine where a function decreases or increases. This approach not only aids in graphical representation but also in optimizing real-world applications where understanding peaks and troughs is necessary.