Problem 35 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ f(x)=x^{5}+x^{3}-2 x+1 $$

Step-by-Step Solution

Answer

Find critical points by solving the derivative equation. Test intervals for derivative sign to find increasing/decreasing intervals.

Step 1 - Find the Derivative

The first step is to find the derivative of the function \( f(x) = x^5 + x^3 - 2x + 1 \). The derivative \( f'(x) \) gives us the rate of change of the function. Using the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \), we calculate:\[ f'(x) = 5x^4 + 3x^2 - 2 \]

Step 2 - Find Critical Points

To find the critical points, we solve the equation \( f'(x) = 0 \). This will help us determine where the function changes from increasing to decreasing or vice versa:\[ 5x^4 + 3x^2 - 2 = 0 \]Factor or use numerical methods to solve for \( x \).

Step 3 - Determine Sign of Derivative

We need to test the intervals between the critical points to determine where the derivative is positive (function is increasing) or negative (function is decreasing). Choose test points in each interval and plug them into \( f'(x) \) to check the sign.

Step 4 - Conclusion on Intervals

Based on the sign test from Step 3:- If \( f'(x) > 0 \), \( f(x) \) is increasing.- If \( f'(x) < 0 \), \( f(x) \) is decreasing.Use this information to specify the ranges of \( x \) for increasing and decreasing intervals.

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus that provide information about the rate of change of a function. When we calculate the derivative of a function, we are essentially finding the slope of the tangent line at any given point on its graph. This tangent line tells us how steep the curve is and in which direction it is moving. For the function given in the exercise, finding its derivative can help us understand how the function behaves and changes.

In this exercise, the function is expressed as a polynomial: \( f(x) = x^5 + x^3 - 2x + 1 \). By taking its derivative, as done in the step-by-step solution, we obtain \( f'(x) = 5x^4 + 3x^2 - 2 \).

This new function \( f'(x) \) represents the slope of the original function \( f(x) \) at any point \( x \). Analyzing this slope over different intervals will reveal where \( f(x) \) is increasing or decreasing.

Identifying Critical Points

Critical points are specific points on the graph of a function where the slope of the tangent is either zero or undefined. These points are crucial because they signify places where the function could switch from increasing to decreasing or vice versa. This behavior helps us locate local maxima, minima, or even points of inflection.To find the critical points, we need to solve the equation \( f'(x) = 0 \), derived from setting just the derivative to zero. For our function's derivative \( f'(x) = 5x^4 + 3x^2 - 2 \), solving \( 5x^4 + 3x^2 - 2 = 0 \) will give us the \( x \) values of the critical points. These are complex polynomial equations often requiring factoring or numerical methods for a solution.Once we have these critical points, they become checkpoints on the chart of the function. They help divide the graph into various intervals for further analysis.

Applying the Power Rule

The power rule is a straightforward and vital tool for finding derivatives of polynomial functions. This rule states that the derivative of \( x^n \) is \( nx^{n-1} \), meaning we multiply the original exponent by the coefficient and subtract one from the exponent.In our exercise, we applied this rule to each term of the polynomial \( f(x) = x^5 + x^3 - 2x + 1 \):

• The derivative of \( x^5 \) is \( 5x^4 \)
• The derivative of \( x^3 \) is \( 3x^2 \)
• The derivative of \( -2x \) is \( -2 \)
• The derivative of a constant \( 1 \) is \( 0 \)

So, the complete derivative using the power rule becomes \( f'(x) = 5x^4 + 3x^2 - 2 \).Understanding and applying the power rule correctly is essential for calculating derivatives quickly and accurately, making it a favorite among calculus students.