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    Math Deficiency - I
    MD-001
    Progress0 / 38 topics
    Topics
    1. Sets: Definition, Representation, and Operations2. Relation and Function: Graphical Transformation of Functions3. Properties of Functions4. Composition and Inverses of Functions5. Domain and Range of Functions6. Maximum and Minimum Values of Functions7. Increasing and Decreasing Functions8. Zeros and Intercepts of Functions9. Piecewise Functions10. Continuity and Discontinuity of Functions11. Polynomials and Rational Functions12. Polynomial Long Division and Synthetic Division13. Solution of Rational Functions14. Absolute Valued Functions and Their Properties15. Asymptotes: Horizontal, Vertical, and Oblique16. Exponential Functions and Their Properties17. Logarithmic Functions and Their Properties18. Systems of Equations: Two Equations and Two Unknowns19. Systems of Equations: Three Equations and Three Unknowns20. Matrix Algebra: Addition, Subtraction, and Multiplication21. Row Operations and Row Echelon Forms22. Augmented Matrices23. Determinant of Matrices: 2x2 and Higher Order24. Cramer's Rule25. Inverse Matrices26. Series and Sequences27. Trigonometry: Angles in Radians and Degrees28. Right Triangle Trigonometry29. Law of Cosines and Sines30. Area of a Triangle31. Graphs of Trigonometric Functions32. Graphs of Inverse Trigonometric Functions33. Basic Trigonometric Identities34. Trigonometric Equations35. General Form of a Conic: Parabolas, Circles, Ellipses, and Hyperbolas36. Degenerate Conics37. Polar and Parametric Equations38. Polar and Rectangular Coordinates
    MD-001›Increasing and Decreasing Functions
    Math Deficiency - ITopic 7 of 38

    Increasing and Decreasing Functions

    14 minread
    2,320words
    Intermediatelevel

    Increasing and Decreasing Functions

    Understanding whether a function is increasing or decreasing is an important concept in calculus and mathematics in general. It provides insight into the behavior of the function, particularly how the output changes as the input (or xxx-value) changes. These concepts are often used in optimization problems, curve sketching, and analyzing the growth or decay of quantities in various fields.


    1. Definition of Increasing and Decreasing Functions

    • Increasing Function: A function f(x)f(x)f(x) is said to be increasing on an interval if, for any two points x1x_1x1​ and x2x_2x2​ within that interval, where x1<x2x_1 < x_2x1​<x2​, it holds that:

      f(x1)≤f(x2)f(x_1) \leq f(x_2)f(x1​)≤f(x2​)

      This means that as xxx increases, the function’s output (or yyy-value) either increases or remains the same.

    • Strictly Increasing Function: If f(x1)<f(x2)f(x_1) < f(x_2)f(x1​)<f(x2​) whenever x1<x2x_1 < x_2x1​<x2​, the function is strictly increasing. In this case, the output always increases as the input increases.

    • Decreasing Function: A function f(x)f(x)f(x) is said to be decreasing on an interval if, for any two points x1x_1x1​ and x2x_2x2​ within that interval, where x1<x2x_1 < x_2x1​<x2​, it holds that:

      f(x1)≥f(x2)f(x_1) \geq f(x_2)f(x1​)≥f(x2​)

      This means that as xxx increases, the function’s output (or yyy-value) either decreases or remains the same.

    • Strictly Decreasing Function: If f(x1)>f(x2)f(x_1) > f(x_2)f(x1​)>f(x2​) whenever x1<x2x_1 < x_2x1​<x2​, the function is strictly decreasing. In this case, the output always decreases as the input increases.


    2. How to Determine Whether a Function is Increasing or Decreasing

    To determine whether a function is increasing or decreasing, we typically use the first derivative of the function.

    Step 1: Find the Derivative of the Function

    The first derivative of a function f(x)f(x)f(x), denoted f′(x)f'(x)f′(x), gives information about the rate of change of the function. Specifically:

    • If f′(x)>0f'(x) > 0f′(x)>0, the function is increasing at that point.
    • If f′(x)<0f'(x) < 0f′(x)<0, the function is decreasing at that point.
    • If f′(x)=0f'(x) = 0f′(x)=0, the function may have a local maximum, minimum, or inflection point.

    Step 2: Analyze the Sign of the Derivative

    • Increasing Function: If the derivative f′(x)>0f'(x) > 0f′(x)>0 on an interval, the function is increasing on that interval.
    • Decreasing Function: If the derivative f′(x)<0f'(x) < 0f′(x)<0 on an interval, the function is decreasing on that interval.

    Step 3: Find Critical Points and Test Intervals

    Critical points occur where f′(x)=0f'(x) = 0f′(x)=0 or f′(x)f'(x)f′(x) does not exist. After finding critical points, you can test the sign of the derivative in the intervals between the critical points. This method is called the First Derivative Test.

    • To determine whether a function is increasing or decreasing on each interval, pick test points within each interval (the intervals determined by the critical points) and substitute them into f′(x)f'(x)f′(x).
      • If f′(x)>0f'(x) > 0f′(x)>0 for a test point in the interval, the function is increasing on that interval.
      • If f′(x)<0f'(x) < 0f′(x)<0 for a test point in the interval, the function is decreasing on that interval.

    3. Example 1: Increasing and Decreasing Intervals of a Function

    Consider the function f(x)=x3−3x2+2xf(x) = x^3 - 3x^2 + 2xf(x)=x3−3x2+2x.

    Step 1: Find the derivative.

    The first derivative of f(x)f(x)f(x) is:

    f′(x)=3x2−6x+2f'(x) = 3x^2 - 6x + 2f′(x)=3x2−6x+2

    Step 2: Find critical points.

    Set the derivative equal to 0 to find the critical points:

    3x2−6x+2=03x^2 - 6x + 2 = 03x2−6x+2=0

    Divide through by 3:

    x2−2x+23=0x^2 - 2x + \frac{2}{3} = 0x2−2x+32​=0

    This equation does not factor nicely, so we solve it using the quadratic formula:

    x=−(−2)±(−2)2−4(1)(2/3)2(1)=2±4−8/32=2±4−2.672x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(2/3)}}{2(1)} = \frac{2 \pm \sqrt{4 - 8/3}}{2} = \frac{2 \pm \sqrt{4 - 2.67}}{2}x=2(1)−(−2)±(−2)2−4(1)(2/3)​​=22±4−8/3​​=22±4−2.67​​

    This results in the critical points x=1.25x = 1.25x=1.25 and x=−1.25x = -1.25x=−1.25.

    Step 3: Analyze the intervals.

    Now, we have critical points at x=1.25x = 1.25x=1.25 and x=−1.25x = -1.25x=−1.25. We divide the real line into intervals based on these points: (−∞,−1.25)(-\infty, -1.25)(−∞,−1.25), (−1.25,1.25)(-1.25, 1.25)(−1.25,1.25), and (1.25,∞)(1.25, \infty)(1.25,∞).

    • For x=−2x = -2x=−2 (a point in the interval (−∞,−1.25)(-\infty, -1.25)(−∞,−1.25)): Substitute into the first derivative:

      f′(−2)=3(−2)2−6(−2)+2=12+12+2=26f'(-2) = 3(-2)^2 - 6(-2) + 2 = 12 + 12 + 2 = 26f′(−2)=3(−2)2−6(−2)+2=12+12+2=26

      Since f′(−2)>0f'(-2) > 0f′(−2)>0, the function is increasing on (−∞,−1.25)(-\infty, -1.25)(−∞,−1.25).

    • For x=0x = 0x=0 (a point in the interval (−1.25,1.25)(-1.25, 1.25)(−1.25,1.25)): Substitute into the first derivative:

      f′(0)=3(0)2−6(0)+2=2f'(0) = 3(0)^2 - 6(0) + 2 = 2f′(0)=3(0)2−6(0)+2=2

      Since f′(0)>0f'(0) > 0f′(0)>0, the function is increasing on (−1.25,1.25)(-1.25, 1.25)(−1.25,1.25).

    • For x=2x = 2x=2 (a point in the interval (1.25,∞)(1.25, \infty)(1.25,∞)): Substitute into the first derivative:

      f′(2)=3(2)2−6(2)+2=12−12+2=2f'(2) = 3(2)^2 - 6(2) + 2 = 12 - 12 + 2 = 2f′(2)=3(2)2−6(2)+2=12−12+2=2

      Since f′(2)>0f'(2) > 0f′(2)>0, the function is increasing on (1.25,∞)(1.25, \infty)(1.25,∞).


    4. Example 2: Determining Decreasing and Increasing Intervals of a Function

    Consider the function f(x)=x2−4x+3f(x) = x^2 - 4x + 3f(x)=x2−4x+3.

    Step 1: Find the derivative.

    The first derivative of f(x)f(x)f(x) is:

    f′(x)=2x−4f'(x) = 2x - 4f′(x)=2x−4

    Step 2: Set the derivative equal to 0 and solve for xxx:

    2x−4=02x - 4 = 02x−4=0 x=2x = 2x=2

    Step 3: Test the intervals.

    We have one critical point at x=2x = 2x=2. We divide the real line into two intervals: (−∞,2)(-\infty, 2)(−∞,2) and (2,∞)(2, \infty)(2,∞).

    • For x=1x = 1x=1 (a point in the interval (−∞,2)(-\infty, 2)(−∞,2)): Substitute into the first derivative:

      f′(1)=2(1)−4=−2f'(1) = 2(1) - 4 = -2f′(1)=2(1)−4=−2

      Since f′(1)<0f'(1) < 0f′(1)<0, the function is decreasing on (−∞,2)(-\infty, 2)(−∞,2).

    • For x=3x = 3x=3 (a point in the interval (2,∞)(2, \infty)(2,∞)): Substitute into the first derivative:

      f′(3)=2(3)−4=2f'(3) = 2(3) - 4 = 2f′(3)=2(3)−4=2

      Since f′(3)>0f'(3) > 0f′(3)>0, the function is increasing on (2,∞)(2, \infty)(2,∞).


    5. Summary of Key Concepts

    Concept Description
    Increasing Function f(x)f(x)f(x) is increasing if f′(x)>0f'(x) > 0f′(x)>0 on an interval.
    Decreasing Function f(x)f(x)f(x) is decreasing if f′(x)<0f'(x) < 0f′(x)<0 on an interval.
    Critical Points Points where f′(x)=0f'(x) = 0f′(x)=0 or f′(x)f'(x)f′(x) does not exist, potential points for local maxima, minima, or inflection.
    First Derivative Test A method to determine whether a function is increasing or decreasing on an interval.
    Strictly Increasing/Decreasing If the derivative is strictly positive (or negative), the function is strictly increasing (or decreasing).

    Conclusion

    • To determine whether a function is increasing or decreasing, find its first derivative f′(x)f'(x)f′(x), and analyze its sign over intervals.
    • Increasing functions have f′(x)>0f'(x) > 0f′(x)>0 over the interval, while decreasing functions have f′(x)<0f'(x) < 0f′(x)<0 over the interval.
    • Use critical points and test intervals to understand the behavior of the function.
    Previous topic 6
    Maximum and Minimum Values of Functions
    Next topic 8
    Zeros and Intercepts of Functions

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