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🚀 Level 3 - Topic 2: Increasing/Decreasing Functions & Extrema 📈📉

Using Derivatives to Find Where Functions Rise, Fall, and Reach Peaks/Valleys

1) Introduction: Analyzing Function Behavior - Increase, Decrease, and Peaks/Valleys

Derivatives provide powerful tools for analyzing the behavior of functions. In this topic, we'll use the first derivative to determine where a function is increasing or decreasing, and to find its extrema (maximum and minimum values). Understanding these aspects is crucial for sketching graphs, optimization problems, and analyzing function behavior in general.

We will cover:

  • Increasing and Decreasing Functions: Using the sign of the first derivative to identify intervals where a function is increasing or decreasing.
  • Critical Numbers and Critical Points: Defining critical numbers and points as candidates for extrema.
  • Local (Relative) Extrema: Finding local maxima and minima using the First Derivative Test.
  • Absolute (Global) Extrema: Finding absolute maxima and minima on closed intervals.
These concepts are essential for understanding the shape and key features of function graphs.

2) Increasing and Decreasing Functions: The First Derivative Test

The sign of the first derivative \( f'(x) \) gives us crucial information about whether a function \( f(x) \) is increasing or decreasing on an interval.

Theorem 6.1: Increasing/Decreasing Test

Let \( f \) be continuous on an interval \( [a, b] \) and differentiable on the open interval \( (a, b) \).

  • If \( f'(x) > 0 \) for all \( x \) in \( (a, b) \), then \( f \) is increasing on \( [a, b] \).
  • If \( f'(x) < 0 \) for all \( x \) in \( (a, b) \), then \( f \) is decreasing on \( [a, b] \).
  • If \( f'(x) = 0 \) for all \( x \) in \( (a, b) \), then \( f \) is constant on \( [a, b] \).

Intuition: A positive derivative means the slope of the tangent line is positive, so the function is going "uphill" (increasing). A negative derivative means the slope is negative, so the function is going "downhill" (decreasing). A zero derivative means the slope is zero (horizontal tangent), indicating the function is momentarily neither increasing nor decreasing (constant at that point).

Examples: Intervals of Increase and Decrease

  1. Example 1: \( f(x) = x^3 - 3x^2 - 9x + 4 \)
    1. Find the first derivative \( f'(x) \).

      \( f'(x) = \frac{d}{dx}[x^3 - 3x^2 - 9x + 4] = 3x^2 - 6x - 9 \).

    2. Find critical numbers by setting \( f'(x) = 0 \) and finding where \( f'(x) \) is undefined.

      \( f'(x) = 3x^2 - 6x - 9 = 0 \Rightarrow 3(x^2 - 2x - 3) = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0 \). So, critical numbers are \( x = 3 \) and \( x = -1 \). \( f'(x) \) is defined for all \( x \).

    3. Create a sign chart for \( f'(x) \) using critical numbers to divide the domain into intervals.

      Critical numbers \( x = -1 \) and \( x = 3 \) divide the real line into intervals \( (-\infty, -1), (-1, 3), (3, \infty) \).

      IntervalTest Value\( f'(x) = 3(x-3)(x+1) \) sign\( f(x) \) behavior
      \( (-\infty, -1) \) (e.g., \( x = -2 \))\( f'(-2) = 3(-5)(-1) = + \)Increasing
      \( (-1, 3) \) (e.g., \( x = 0 \))\( f'(0) = 3(-3)(1) = - \)Decreasing
      \( (3, \infty) \) (e.g., \( x = 4 \))\( f'(4) = 3(1)(5) = + \)Increasing
    4. State the intervals of increase and decrease.

      \( f(x) \) is increasing on \( (-\infty, -1] \) and \( [3, \infty) \). \( f(x) \) is decreasing on \( [-1, 3] \).

  2. Example 2: \( g(x) = \frac{x}{x^2 + 1} \)
    1. Find the first derivative \( g'(x) \) using the Quotient Rule.

      \( g'(x) = \frac{d}{dx}\left[\frac{x}{x^2 + 1}\right] = \frac{(1)(x^2 + 1) - x(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \).

    2. Find critical numbers by setting \( g'(x) = 0 \).

      \( g'(x) = \frac{1 - x^2}{(x^2 + 1)^2} = 0 \Rightarrow 1 - x^2 = 0 \) (numerator must be zero) \( \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \). The denominator \( (x^2 + 1)^2 \) is never zero and \( g'(x) \) is defined for all \( x \).

    3. Create a sign chart for \( g'(x) \) using critical numbers \( x = -1, 1 \).

      Intervals are \( (-\infty, -1), (-1, 1), (1, \infty) \).

      IntervalTest Value\( g'(x) = \frac{1-x^2}{(x^2+1)^2} \) sign\( g(x) \) behavior
      \( (-\infty, -1) \) (e.g., \( x = -2 \))\( g'(-2) = \frac{1-(-2)^2}{((-) +)^2} = \frac{-3}{+} = - \)Decreasing
      \( (-1, 1) \) (e.g., \( x = 0 \))\( g'(0) = \frac{1-0^2}{(+)^2} = \frac{1}{+} = + \)Increasing
      \( (1, \infty) \) (e.g., \( x = 2 \))\( g'(2) = \frac{1-(2)^2}{(+)^2} = \frac{-3}{+} = - \)Decreasing
    4. State the intervals of increase and decrease.

      \( g(x) \) is increasing on \( [-1, 1] \). \( g(x) \) is decreasing on \( (-\infty, -1] \) and \( [1, \infty) \).

3) Critical Numbers and Local Extrema

Critical numbers are crucial points for finding local extrema.

Definition 6.1: Critical Number and Critical Point

  • A critical number of a function \( f \) is a number \( c \) in the domain of \( f \) such that either \( f'(c) = 0 \) or \( f'(c) \) does not exist.
  • A critical point is a point \( (c, f(c)) \) on the graph of \( f \) where \( c \) is a critical number.

Local extrema (local maxima and local minima) occur at critical points (or endpoints of the domain).

Definition 6.2: Local Maximum and Local Minimum

  • A function \( f \) has a local maximum at \( x = c \) if \( f(c) \geq f(x) \) when \( x \) is near \( c \) (for all \( x \) in some open interval containing \( c \)). Visually, it's a "peak" in the nearby region.
  • A function \( f \) has a local minimum at \( x = c \) if \( f(c) \leq f(x) \) when \( x \) is near \( c \) (for all \( x \) in some open interval containing \( c \)). Visually, it's a "valley" in the nearby region.

4) The First Derivative Test for Local Extrema

First Derivative Test

Suppose \( c \) is a critical number of a continuous function \( f \).

  • If \( f'(x) \) changes from positive to negative at \( x = c \) (as \( x \) increases through \( c \)), then \( f \) has a local maximum at \( c \).
  • If \( f'(x) \) changes from negative to positive at \( x = c \) (as \( x \) increases through \( c \)), then \( f \) has a local minimum at \( c \).
  • If \( f'(x) \) does not change sign at \( x = c \) (i.e., \( f'(x) \) is positive on both sides of \( c \) or negative on both sides), then \( f \) has no local extremum at \( c \) (neither a local max nor a local min).

Using the sign chart: We can use the sign chart for \( f'(x) \) we created for increasing/decreasing intervals to identify local extrema.

  • Local Max: Function increases then decreases around \( x = c \) ( \( f'(x) \) changes from + to - ).
  • Local Min: Function decreases then increases around \( x = c \) ( \( f'(x) \) changes from - to + ).
  • No Local Extremum: Function keeps increasing or keeps decreasing around \( x = c \) ( \( f'(x) \) sign does not change).

Examples: Finding Local Extrema using First Derivative Test

  1. Example 1 (revisited): \( f(x) = x^3 - 3x^2 - 9x + 4 \)

    From our increasing/decreasing analysis:

    • At \( x = -1 \): \( f'(x) \) changes from positive to negative (increasing to decreasing). Thus, \( f \) has a local maximum at \( x = -1 \). Local maximum value is \( f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 4 = -1 - 3 + 9 + 4 = 9 \). Local maximum point is \( (-1, 9) \).
    • At \( x = 3 \): \( f'(x) \) changes from negative to positive (decreasing to increasing). Thus, \( f \) has a local minimum at \( x = 3 \). Local minimum value is \( f(3) = (3)^3 - 3(3)^2 - 9(3) + 4 = 27 - 27 - 27 + 4 = -23 \). Local minimum point is \( (3, -23) \).
  2. Example 2 (revisited): \( g(x) = \frac{x}{x^2 + 1} \)

    From our increasing/decreasing analysis:

    • At \( x = -1 \): \( g'(x) \) changes from negative to positive (decreasing to increasing). Thus, \( g \) has a local minimum at \( x = -1 \). Local minimum value is \( g(-1) = \frac{-1}{(-1)^2 + 1} = \frac{-1}{2} = -\frac{1}{2} \). Local minimum point is \( (-1, -\frac{1}{2}) \).
    • At \( x = 1 \): \( g'(x) \) changes from positive to negative (increasing to decreasing). Thus, \( g \) has a local maximum at \( x = 1 \). Local maximum value is \( g(1) = \frac{1}{(1)^2 + 1} = \frac{1}{2} \). Local maximum point is \( (1, \frac{1}{2}) \).
  3. Example 3: \( h(x) = x^4 \)
    1. Find \( h'(x) \). \( h'(x) = 4x^3 \).
    2. Critical numbers: \( h'(x) = 4x^3 = 0 \Rightarrow x = 0 \).
    3. Sign chart for \( h'(x) \):
      IntervalTest Value\( h'(x) = 4x^3 \) sign\( h(x) \) behavior
      \( (-\infty, 0) \) (e.g., \( x = -1 \))\( h'(-1) = 4(-1)^3 = - \)Decreasing
      \( (0, \infty) \) (e.g., \( x = 1 \))\( h'(1) = 4(1)^3 = + \)Increasing
    4. First Derivative Test: At \( x = 0 \), \( h'(x) \) changes from negative to positive. Thus, \( h \) has a local minimum at \( x = 0 \). Local minimum value is \( h(0) = 0^4 = 0 \). Local minimum point is \( (0, 0) \).
  4. Example 4: \( k(x) = x^3 \)
    1. Find \( k'(x) \). \( k'(x) = 3x^2 \).
    2. Critical numbers: \( k'(x) = 3x^2 = 0 \Rightarrow x = 0 \).
    3. Sign chart for \( k'(x) \):
      IntervalTest Value\( k'(x) = 3x^2 \) sign\( k(x) \) behavior
      \( (-\infty, 0) \) (e.g., \( x = -1 \))\( k'(-1) = 3(-1)^2 = + \)Increasing
      \( (0, \infty) \) (e.g., \( x = 1 \))\( k'(1) = 3(1)^2 = + \)Increasing
    4. First Derivative Test: At \( x = 0 \), \( k'(x) \) does not change sign (it's positive on both sides, or more accurately non-negative on both sides). Thus, \( k \) has no local extremum at \( x = 0 \). It is neither a local max nor a local min. The function is increasing through \( x = 0 \), even though \( f'(0) = 0 \).

5) Absolute (Global) Extrema on Closed Intervals

Absolute extrema are the true maximum and minimum values of a function over its entire domain or a specified interval. On a closed interval \( [a, b] \), a continuous function \( f \) is guaranteed to have both an absolute maximum and an absolute minimum value (Extreme Value Theorem). These absolute extrema can occur at:

  • Critical points within the interval \( (a, b) \) (where \( f'(x) = 0 \) or \( f'(x) \) DNE).
  • Endpoints of the interval \( x = a \) and \( x = b \).

Procedure to Find Absolute Extrema on a Closed Interval \( [a, b] \)

  1. Find all critical numbers of \( f \) in the interval \( (a, b) \). These are values \( c \) where \( a < c < b \) and \( f'(c) = 0 \) or \( f'(c) \) is undefined.
  2. Evaluate \( f \) at all critical numbers found in step 1.
  3. Evaluate \( f \) at the endpoints of the interval, \( x = a \) and \( x = b \).
  4. Compare the function values from steps 2 and 3.
    • The largest value is the absolute maximum value on \( [a, b] \).
    • The smallest value is the absolute minimum value on \( [a, b] \).

Examples: Finding Absolute Extrema on Closed Intervals

  1. Example 1: Find the absolute extrema of \( f(x) = x^3 - 3x^2 - 9x + 4 \) on the interval \( [0, 4] \).
    1. Find critical numbers in \( (0, 4) \). We found critical numbers \( x = -1 \) and \( x = 3 \) earlier. Only \( x = 3 \) is in the interval \( (0, 4) \).
    2. Evaluate \( f \) at critical number(s) in \( (0, 4) \). \( f(3) = -23 \) (calculated before).
    3. Evaluate \( f \) at endpoints \( x = 0 \) and \( x = 4 \).

      \( f(0) = (0)^3 - 3(0)^2 - 9(0) + 4 = 4 \).

      \( f(4) = (4)^3 - 3(4)^2 - 9(4) + 4 = 64 - 48 - 36 + 4 = -16 \).

    4. Compare values: \( f(3) = -23, f(0) = 4, f(4) = -16 \).

      Largest value is 4 (at \( x = 0 \)), smallest value is -23 (at \( x = 3 \)).

      Absolute maximum value is 4, absolute maximum occurs at \( x = 0 \). Absolute minimum value is -23, absolute minimum occurs at \( x = 3 \).

  2. Example 2: Find absolute extrema of \( g(x) = x^2 - 4\sqrt{x} \) on \( [0, 4] \).
    1. Find \( g'(x) \).

      \( g(x) = x^2 - 4x^{1/2} \Rightarrow g'(x) = 2x - 4 \cdot \frac{1}{2}x^{-1/2} = 2x - 2x^{-1/2} = 2x - \frac{2}{\sqrt{x}} \).

    2. Find critical numbers in \( (0, 4) \).

      Set \( g'(x) = 2x - \frac{2}{\sqrt{x}} = 0 \Rightarrow 2x = \frac{2}{\sqrt{x}} \Rightarrow x = \frac{1}{\sqrt{x}} \Rightarrow x\sqrt{x} = 1 \Rightarrow x^{3/2} = 1 \Rightarrow x = 1^{2/3} = 1 \). So \( x = 1 \) is a critical number. Also, \( g'(x) = 2x - \frac{2}{\sqrt{x}} \) is undefined at \( x = 0 \), but \( x = 0 \) is an endpoint, not in \( (0, 4) \). So, only critical number in \( (0, 4) \) is \( x = 1 \).

    3. Evaluate \( g \) at critical number(s) in \( (0, 4) \). \( g(1) = (1)^2 - 4\sqrt{1} = 1 - 4 = -3 \).
    4. Evaluate \( g \) at endpoints \( x = 0 \) and \( x = 4 \).

      \( g(0) = (0)^2 - 4\sqrt{0} = 0 \).

      \( g(4) = (4)^2 - 4\sqrt{4} = 16 - 4(2) = 16 - 8 = 8 \).

    5. Compare values: \( g(1) = -3, g(0) = 0, g(4) = 8 \).

      Largest value is 8 (at \( x = 4 \)), smallest value is -3 (at \( x = 1 \)).

      Absolute maximum value is 8, absolute maximum occurs at \( x = 4 \). Absolute minimum value is -3, absolute minimum occurs at \( x = 1 \).

6) Practice Questions - Increasing/Decreasing Functions and Extrema

Test your understanding of increasing/decreasing functions and finding extrema.

Fundamental Practice Questions

Instructions: For each function, find the intervals where it is increasing and decreasing, and find all local extrema.

  1. Q1. \( f(x) = x^2 - 6x + 5 \)
  2. Q2. \( g(x) = 2x^3 + 3x^2 - 12x \)
  3. Q3. \( h(x) = x^4 - 4x^3 + 4x^2 \)
  4. Q4. \( y = \frac{x}{x - 1} \)
  5. Q5. \( f(x) = \sqrt{x^2 + 1} \)
  6. Q6. \( g(x) = \sin(x) + \cos(x) \) on \( [0, 2\pi] \)
  7. Q7. \( h(x) = x e^{-x} \)
  8. Q8. \( y = x - 2\sin(x) \) on \( [0, 2\pi] \)
  9. Q9. \( f(x) = \frac{x^2}{x^2 + 4} \)
  10. Q10. \( g(x) = x^{1/3}(x + 4) \)

Challenging Practice Questions

Instructions: Find the absolute maximum and minimum values of each function on the given closed interval.

  1. Q1. \( f(x) = x^3 - 6x^2 + 5 \) on \( [-3, 5] \)
  2. Q2. \( g(x) = x^4 - 2x^2 + 3 \) on \( [-2, 2] \)
  3. Q3. \( h(x) = \frac{x}{x^2 + 1} \) on \( [-2, 2] \)
  4. Q4. \( y = x - \ln(x) \) on \( [0.5, 3] \)
  5. Q5. \( f(x) = x^{2/3} - x \) on \( [-1, 3] \)

7) Summary & Cheat Sheet - Increasing/Decreasing Functions and Extrema

Concise summary of key concepts and procedures.

7.1) Increasing/Decreasing Test

  • If \( f'(x) > 0 \) on an interval, \( f \) is increasing.
  • If \( f'(x) < 0 \) on an interval, \( f \) is decreasing.

7.2) First Derivative Test for Local Extrema

  • Local Max at \( c \) if \( f'(x) \) changes from + to - at \( x = c \).
  • Local Min at \( c \) if \( f'(x) \) changes from - to + at \( x = c \).
  • No Local Extremum if \( f'(x) \) does not change sign at \( x = c \).

7.3) Finding Absolute Extrema on a Closed Interval \( [a, b] \)

  1. Find critical numbers in \( (a, b) \).
  2. Evaluate \( f \) at critical numbers.
  3. Evaluate \( f \) at endpoints \( a \) and \( b \).
  4. Compare values – largest is absolute max, smallest is absolute min.

Excellent work! You've now learned how to use the first derivative to analyze where functions increase, decrease, and find local and absolute extrema. These are fundamental tools for understanding function behavior and optimization. Next, we'll explore how the second derivative adds even more insights!

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