Problem 31 - Calculus with Concepts in Calculus

Question

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

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, \infty)\) and not decreasing anywhere.

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, we first find the derivative of the function. The function given is \( f(x) = x^3 - x^2 + x - 1 \).The derivative, \( f'(x) \), is found by differentiating each term separately:\( f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(x^2) + \frac{d}{dx}(x) - \frac{d}{dx}(1) = 3x^2 - 2x + 1 \).

Step 2: Determine Critical Points

To find the intervals where the function changes from increasing to decreasing or vice versa, we solve \( f'(x) = 0 \).Set the derivative equal to zero: \( 3x^2 - 2x + 1 = 0 \).Use the quadratic formula to solve \( ax^2 + bx + c = 0 \), where \( a = 3 \), \( b = -2 \), and \( c = 1 \):\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).Substitute in the values: \( x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \).Calculate inside the square root: \( \sqrt{4 - 12} = \sqrt{-8} \), which is imaginary.Since the discriminant (4 - 12) is negative, there are no real roots, meaning \( f'(x) \) doesn't equal zero for any real number.

Step 3: Test Intervals for Increasing or Decreasing Behavior

Since there are no real critical points from Step 2, we analyze the sign of \( f'(x) = 3x^2 - 2x + 1 \) to determine monotonicity.Consider that \( 3x^2 - 2x + 1 \) is a quadratic that always opens upwards (\( a = 3 > 0 \)), and because it does not intersect the x-axis, it is always positive.Thus, \( f'(x) > 0 \) for all real numbers, indicating that \( f(x) \) is increasing on its entire domain, \( (-\infty, \infty) \).

Key Concepts

Critical Points

Critical points of a function are essential in determining where a function may change its behavior from increasing to decreasing or vice versa. To find these points, we first need to compute the derivative of the function. For instance, in our exercise, the function is given as \( f(x) = x^3 - x^2 + x - 1 \). The derivative, \( f'(x) \), is \( 3x^2 - 2x + 1 \).
Critical points occur when the derivative is equal to zero, as it indicates potential changes in the direction of the graph (think of a hill or valley). Therefore, we solve \( 3x^2 - 2x + 1 = 0 \) to locate these points.
However, in our case, the roots of this equation are imaginary since the discriminant \( 4 - 12 = -8 \) is negative. This means there are no real x-values for which \( f'(x) = 0 \), indicating that there are no critical points within the real number system for this function.

Monotonicity

A function is described as monotonic if it is either entirely non-increasing or non-decreasing throughout its domain. This can be determined by analyzing the sign of its derivative. If the derivative is consistently positive over an interval, the function is increasing in that interval. Similarly, if the derivative is consistently negative, the function is decreasing.
For the given function \( f(x) = x^3 - x^2 + x - 1 \), we have already determined that the derivative is \( f'(x) = 3x^2 - 2x + 1 \). Notice that the expression \( 3x^2 - 2x + 1 \) represents a quadratic function that always opens upwards because the leading coefficient (3) is positive, and it does not have any real roots. Consequently, \( f'(x) \) is always positive, making the function consistently increasing on its entire domain.
Thus, the function \( f(x) \) is monotonic, specifically monotonically increasing, across all real numbers \( (-\infty, \infty) \).

Intervals of Increase and Decrease

Finding the intervals in which a function is increasing or decreasing forms a significant aspect of understanding its overall behavior. These intervals help to establish where a function is ascending or descending as we move from left to right along the x-axis.
Given the function \( f(x) = x^3 - x^2 + x - 1 \), and our evaluation of its derivative \( f'(x) = 3x^2 - 2x + 1 \), we found that \( f'(x) \) does not cross below zero at any point due to no real critical points.
Consequently, since \( f'(x) \) is always positive, \( f(x) \) does not experience any intervals of decrease.

• The function is increasing on the entire domain \( (-\infty, \infty) \), as the derivative does not change signs.

This consistency means there are no separate intervals where the function switches from increasing to decreasing or vice versa.