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Chapter 3: Problem 23

23- 30 . A function \(f\) is given. (a) Use a graphing device to draw the graph of \(f .\) (b) State approximately the intervals on which \(f\) is increasing and on which \(f\) is decreasing. $$ f(x)=x^{2}-5 x $$

Short Answer

Expert verified
The function is decreasing on \((-\infty, \frac{5}{2})\) and increasing on \((\frac{5}{2}, \infty)\).

Step by step solution

01

Analyze the Function

The given function is a quadratic, \( f(x) = x^2 - 5x \). It has a standard parabola shape that opens upwards because the coefficient of \( x^2 \) is positive.
02

Calculate the Vertex

To find the vertex of the parabola, we can use the formula for the vertex of a quadratic function, \( x = \frac{-b}{2a} \), where \( a = 1 \) and \( b = -5 \). Substituting these values gives us \( x = \frac{-(-5)}{2(1)} = \frac{5}{2} \). So, the vertex is at \( x = \frac{5}{2} \).
03

Determine the Sign of the Derivative

The behavior of the function (increasing or decreasing) can be determined by examining the sign of its derivative. The derivative of \( f(x) = x^2 - 5x \) is \( f'(x) = 2x - 5 \).
04

Find Critical Points

To find critical points, we set \( f'(x) = 0 \). Solving \( 2x - 5 = 0 \) yields \( x = \frac{5}{2} \). This is the critical point where the function's behavior changes.
05

Test Intervals Around the Critical Point

The critical point \( x = \frac{5}{2} \) divides the x-axis into two intervals: \( (-\infty, \frac{5}{2}) \) and \( (\frac{5}{2}, \infty) \). Choose test points in each interval (e.g., \( x = 0 \) for the first and \( x = 3 \) for the second) to determine the sign of the derivative. For \( x = 0 \), \( f'(0) = 2(0) - 5 = -5 \), indicating decreasing. For \( x = 3 \), \( f'(3) = 2(3) - 5 = 1 \), indicating increasing.
06

Graph the Function

Using a graphing device (like a graphing calculator or software), plot the function \( f(x) = x^2 - 5x \). The graph is a parabola opening upwards with vertex at \( x = \frac{5}{2} \). The graph visually confirms the intervals found for increasing and decreasing.
07

Conclude Increasing and Decreasing Intervals

Based on the analysis and graph, the function is decreasing on \( (-\infty, \frac{5}{2}) \) and increasing on \( (\frac{5}{2}, \infty) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabola
In the study of quadratic functions, the graph is typically a shape known as a parabola. For a function given by \( f(x) = ax^2 + bx + c \), if \( a \) is positive, the parabola will open upwards, like a smiling face. Conversely, if \( a \) is negative, it opens downwards, resembling a frown.
  • For the given function \( f(x) = x^2 - 5x \), the parabola opens upwards because the coefficient of \( x^2 \) is 1, which is positive.
  • The general shape helps us quickly identify some properties, such as whether it reaches a minimum or maximum point.
Understanding the basic shape of a parabola allows us to predict the behavior of the function at various points across its domain.
Vertex
The vertex of a parabola is a crucial point as it can be either a minimum or a maximum for the function. It signifies the point where the direction of the graph changes.
  • The vertex of a parabola given by \( f(x) = ax^2 + bx + c \) can be calculated using the formula \( x = \frac{-b}{2a} \).
  • For \( f(x) = x^2 - 5x \), substitute \( a = 1 \) and \( b = -5 \) into the formula, yielding \( x = \frac{5}{2} \).
This calculation shows that the vertex occurs at \( x = \frac{5}{2} \). A more precise definition of the vertex could be found if we substituted \( x = \frac{5}{2} \) back into the function to find the corresponding \( y ext{-value} \). In this case, the vertex of the parabola is \( (\frac{5}{2}, -\frac{25}{4}) \). The vertex serves as the turning point of the graph.
Derivative
The derivative of a function provides insight into its rate of change. By finding the derivative of a quadratic function, we can determine where it is increasing or decreasing.
  • For \( f(x) = x^2 - 5x \), its derivative is \( f'(x) = 2x - 5 \).
  • The derivative helps find the slope of the tangent at any point \( x \). If \( f'(x) > 0 \), the function is increasing at \( x \); if \( f'(x) < 0 \), it is decreasing.
The point where the derivative equals zero, i.e., \( f'(x) = 0 \), is known as a critical point. These points help determine where the function changes behavior, from increasing to decreasing, or vice versa.
Increasing Intervals
An interval in which a function is increasing means that as \( x \) moves through this interval, \( f(x) \) grows larger. The derivative of the function is key to identifying these intervals.
  • If \( f'(x) > 0 \) for every \( x \) in an interval, then the function is increasing in that interval.
  • In the function \( f(x) = x^2 - 5x \), since \( f'(x) = 2x - 5 \), it is greater than zero when \( x > \frac{5}{2} \).
Therefore, the function is increasing on the interval \((\frac{5}{2}, \infty) \). This means after \( x = \frac{5}{2} \), the values of \( f(x) \) keep rising as \( x \) continues to increase.
Decreasing Intervals
If a function is decreasing over an interval, it means that as \( x \) progresses through this interval, \( f(x) \) becomes smaller.
  • The function's derivative, \( f'(x) \), helps support this identification. A function decreases where its derivative is less than zero.
  • For \( f(x) = x^2 - 5x \), since \( f'(x) = 2x - 5 \), it is less than zero when \( x < \frac{5}{2} \).
Hence, the function decreases on the interval \((-\infty, \frac{5}{2}) \). Knowing where a function decreases can greatly assist in graphing and understanding the behavior of quadratic functions across different domains.

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