Problem 97 - Calculus and its applications

Question

Find the interval(s) for which \(f^{\prime}(x)\) is positive. Find the points on the graph of $$ y=x^{4}-\frac{4}{3} x^{2}-4 $$

Step-by-Step Solution

Answer

The derivative \( f'(x) \) is positive on \((-\sqrt{\frac{2}{3}}, 0)\) and \((\sqrt{\frac{2}{3}}, \infty)\).

Step 1: Find the Derivative

First, we need to find the derivative of the given function \( y = x^4 - \frac{4}{3}x^2 - 4 \). The derivative, \( f'(x) \), of the function is found by differentiating each term: \[ f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}\left(\frac{4}{3}x^2\right) - \frac{d}{dx}(4) \]This gives us: \[ f'(x) = 4x^3 - \frac{8}{3}x \].

Step 2: Solve for Critical Points

To determine where \( f'(x) \) is positive, we first find where it is zero (critical points) by solving:\[ 4x^3 - \frac{8}{3}x = 0 \].Factor out the common term, \( 4x \):\[ 4x(x^2 - \frac{2}{3}) = 0 \].Setting each factor to zero gives:\( x = 0 \) and \( x^2 = \frac{2}{3} \).Solving \( x^2 = \frac{2}{3} \), we get \( x = \pm \sqrt{\frac{2}{3}} \). So, the critical points are \( x = 0, \sqrt{\frac{2}{3}}, -\sqrt{\frac{2}{3}} \).

Step 3: Test Intervals Around Critical Points

The critical points \( x = -\sqrt{\frac{2}{3}}, 0, \sqrt{\frac{2}{3}} \) divide the number line into four intervals:1. \( (-\infty, -\sqrt{\frac{2}{3}}) \)2. \( (-\sqrt{\frac{2}{3}}, 0) \)3. \( (0, \sqrt{\frac{2}{3}}) \)4. \( (\sqrt{\frac{2}{3}}, \infty) \)Choose test points from each interval to test the sign of \( f'(x) \):- For \,\[(-\infty, -\sqrt{\frac{2}{3}})\],\ try \,\[x = -1\]: \( f'(-1) = 4(-1)^3 - \frac{8}{3}(-1) = -4 + \frac{8}{3} = -\frac{4}{3} \) (negative)- For \,\[(-\sqrt{\frac{2}{3}}, 0)\],\ try \,\[x = -0.5\]: \( f'(-0.5) = 4(-0.5)^3 - \frac{8}{3}(-0.5) = -1 + \frac{4}{3} = \frac{1}{3} \) (positive)- For \,\[(0, \sqrt{\frac{2}{3}})\],\ try \,\[x = 0.5\]: \( f'(0.5) = 4(0.5)^3 - \frac{8}{3}(0.5) = 1 - \frac{4}{3} = -\frac{1}{3} \) (negative)- For \,\[(\sqrt{\frac{2}{3}}, \infty)\],\ try \,\[x = 1\]: \( f'(1) = 4(1)^3 - \frac{8}{3}(1) = 4 - \frac{8}{3} = \frac{4}{3} \) (positive)

Step 4: Identify Positive Intervals

From the results of the test points, we see that \( f'(x) > 0 \) in the intervals:\[ (-\sqrt{\frac{2}{3}}, 0) \] and \[ (\sqrt{\frac{2}{3}}, \infty) \].

Key Concepts

Critical Points

Finding critical points is a crucial part of analyzing a function. They help us understand the behavior of the function by identifying where the function's slope changes direction—essentially where it might have a peak or a valley. To find these points, one must first compute the derivative of the function, as these points occur where the derivative is zero or undefined.
In this exercise, the derivative is calculated to be \( f'(x) = 4x^3 - \frac{8}{3}x \). Setting the derivative equal to zero gives us the critical points. Factoring, we have \( 4x(x^2 - \frac{2}{3}) = 0 \), which leads to \( x = 0, \sqrt{\frac{2}{3}}, -\sqrt{\frac{2}{3}} \).
These values of \( x \) are where the function changes direction, making them important markers on the graph.

Sign of Derivative

The sign of the derivative tells us whether the function is increasing or decreasing. When the derivative, \( f'(x) \), is positive, our function is rising; when it's negative, our function is falling.
In our problem, the solution involves checking the sign of \( f'(x) \) around the critical points. This determines whether \( f'(x) \) is positive or negative in those regions, thereby revealing where the function increases or decreases.
This method allows us to sketch the graph more accurately and understand how the function behaves between the critical points.

Interval Testing

Interval testing involves selecting specific values (or test points) from defined intervals on the number line. This step allows us to determine the sign of the derivative in each interval. It is crucial in identifying intervals where the derivative is consistently positive or negative.
The critical points \( x = -\sqrt{\frac{2}{3}}, 0, \sqrt{\frac{2}{3}} \) divide the number line into four intervals: \((-\infty, -\sqrt{\frac{2}{3}})\), \((-\sqrt{\frac{2}{3}}, 0)\), \((0, \sqrt{\frac{2}{3}})\), and \((\sqrt{\frac{2}{3}}, \infty)\).
You pick a test point from each of these intervals to substitute into \( f'(x) \).

• For \((-\infty, -\sqrt{\frac{2}{3}})\), try \( x = -1 \). The derivative is negative.
• For \((-\sqrt{\frac{2}{3}}, 0)\), try \( x = -0.5 \). The derivative is positive.
• For \((0, \sqrt{\frac{2}{3}})\), try \( x = 0.5 \). The derivative is negative.
• For \((\sqrt{\frac{2}{3}}, \infty)\), try \( x = 1 \). The derivative is positive.

Through this evaluation, you can conclude the function is increasing wherever the derivative is positive, specifically in \((-\sqrt{\frac{2}{3}}, 0)\) and \((\sqrt{\frac{2}{3}}, \infty)\).