Problem 33 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ f(x)=x^{4}-2 x^{3}+1 $$

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, 0)\) and \((0, \frac{3}{2})\) and increasing on \((\frac{3}{2}, \infty)\).

Step 1: Differentiate the Function

Find the first derivative of the function, as it helps to determine where the function is increasing or decreasing. The derivative, denoted as \( f'(x) \), is given by applying the power rule:\[f'(x) = \frac{d}{dx}(x^4 - 2x^3 + 1) = 4x^3 - 6x^2\]

Step 2: Find Critical Points

Set the derivative \( f'(x) \) equal to zero and solve for \( x \) to find the critical points.\[4x^3 - 6x^2 = 0\]Factor the equation:\[2x^2(2x - 3) = 0\]This gives solutions \( x = 0 \) and \( x = \frac{3}{2} \). These are the critical points.

Step 3: Test Intervals Around Critical Points

Determine intervals around the critical points. Choose test points in each interval to determine if \( f'(x) \) is positive (increasing function) or negative (decreasing function):- Interval \((-\infty, 0)\): Choose \( x = -1 \), \( f'(-1) = 4(-1)^3 - 6(-1)^2 = -4 - 6 = -10 \) (decreasing).- Interval \((0, \frac{3}{2})\): Choose \( x = 1 \), \( f'(1) = 4(1)^3 - 6(1)^2 = 4 - 6 = -2 \) (decreasing).- Interval \((\frac{3}{2}, \infty)\): Choose \( x = 2 \), \( f'(2) = 4(2)^3 - 6(2)^2 = 32 - 24 = 8 \) (increasing).

Step 4: Conclusion

Based on the test points:- The function is decreasing on \((-\infty, 0)\) and \((0, \frac{3}{2})\).- The function is increasing on \((\frac{3}{2}, \infty)\).

Key Concepts

Increasing and Decreasing Intervals

To understand where a function is increasing or decreasing, we look at its behavior between critical points. These are special points on the function where the slope of the tangent, or the first derivative, is zero.
The concept of increasing and decreasing intervals can be explained by the sign of the first derivative over certain intervals.

• If the first derivative is positive ( \( f'(x) > 0 \) ), the function is increasing in that interval.
• If the first derivative is negative ( \( f'(x) < 0 \) ), the function is decreasing in that interval.

By determining the sign of the derivative across different intervals, especially between critical points, we can precisely describe the intervals where a function rises or falls. In our exercise, the function is found to be decreasing on \((-\infty, 0)\) and \((0, \frac{3}{2})\), and increasing on \((\frac{3}{2}, \infty)\). Understanding these intervals helps in sketching the graph and in solving further advanced problems.

Critical Points

Critical points play a vital role in calculus as they help find the turning points of a function, where the function changes its increasing or decreasing behavior.
These points occur where the first derivative of a function equals zero or is undefined.Critical points help us in:

• Identifying maxima, minima, or saddle points of functions.
• Understanding where the slope shifts between negative to positive or vice versa.

In our specific exercise, solving the equation\[ 4x^3 - 6x^2 = 0 \]using factoring, we identify the critical points as \( x = 0 \) and \( x = \frac{3}{2} \). These points divide the function into intervals where we can analyze the sign of the first derivative to find where the function is increasing or decreasing.

First Derivative

The first derivative of a function is the cornerstone of understanding its rate of change and slope at any given point. In practical terms, finding the first derivative is like finding the formula for the slope of a curve.When solving problems like our exercise:

• Calculate the derivative using differentiation rules like the power rule.
• The first derivative \( f'(x) \) tells us how the function \( f(x) \) behaves.

In the problem, by finding the first derivative as:\[ f'(x) = 4x^3 - 6x^2 \]we gain insight into the function's behavior by analyzing where this expression is zero (finding critical points) or positive and negative (determining increasing or decreasing intervals). Therefore, the first derivative is crucial in exploring and understanding the dynamics of the function in calculus.