Calculus: Maxima and Minima PDF

## Our Goal: Learn about f(x) from df/dx. We ask two quick questions: 1. If df/dx is positive, what does that say about f? 2. If the slope is negative, how is that reflected in the function? Then, the critical question: **How do you identify a maximum or a minimum?** Normal answer: The slope...

## Our Goal: Learn about f(x) from df/dx. We ask two quick questions: 1. If df/dx is positive, what does that say about f? 2. If the slope is negative, how is that reflected in the function? Then, the critical question: **How do you identify a maximum or a minimum?** Normal answer: The slope is _zero_. ### Reaching df/dx = 0 This may be the most important application of calculus. #### The Easy Questions Suppose df/dx is positive for every x between a and b. All tangent lines slope upwards. The function f(x) is increasing as x goes from a to b. - If df/dx > 0 then f(x) is increasing. - If df/dx < 0 then f(x) is decreasing. #### Defining Increasing and Decreasing Look at any two points x < X. "Increasing" requires f(x) < f(X). "Decreasing" requires f(x) > f(X). A positive slope does not mean a positive function. The function itself can be positive or negative. ### Example 1: The Function f(x) = x² - 2x This function has a slope of 2x - 2. This slope is positive when x > 1 and negative when x < 1. The function increases after x = 1 and decreases before x = 1. ## 3.2 Maximum and Minimum Problems **Figure 3.3:** Slopes are +++, Slope is --- so f is up-down-up-down-up. **Description:** * The figure shows two parabolas, each with a different equation. The top parabola is x²−2x+5 and the bottom one is x²−2x+1. They both have the same slope 2x - 2. * The figure also plots the slope, which is 2x - 2. * By comparing the slope with the graph of the parabolas, we can see how the slope determines whether the function is increasing or decreasing - the parabola is increasing when the slope is positive and decreasing when the slope is negative. We say that without computing f(x) at any point! The parabola in Figure 3.3 goes down to its minimum at x = 1 and up again. ### Example 2: The Function x² - 2x + 5 This function has the same slope (2x - 2) as the previous function, but its graph is shifted up by 5. - All functions with slope 2x - 2 are parabolas of the form x² - 2x + C, shifted up or down according to C. - Some parabolas cross the x-axis (those crossings are solutions to f(x) = 0). - Other parabolas stay entirely above the x-axis. - The special parabola x² - 2x + 1 = (x - 1)² grazes the x-axis at x = 1. It has a "double zero" where f(x) = df/dx = 0. ### Example 3: The Function df/dx = (x−1)(x−2)(x−3)(x−4) This slope is positive beyond x = 4 and up to x = 1 (df/dx = 24 at x = 0). The slope is positive again between 2 and 3. At x = 1, 2, 3, 4 this slope is zero and f(x) changes direction. - Here f(x) is a 5th-degree polynomial, because f'(x) is fourth-degree. - The graph of f goes up-down-up-down-up. - It might cross the x-axis five times. However, it must cross it at least once. - When complex numbers are allowed, every 5th degree polynomial has five roots. **Important:** * "Positive slope implies increasing function" is obvious - perhaps it is. However, we still have to deduce f(X) > f(x) at pairs of points. This is a “local to global” question. * The Mean Value Theorem will help us solve this. * We could also wait for the Fundamental Theorem of Calculus: The difference f(X) - f(x) equals the area under the graph of df/dx. That area is positive, so f(X) exceeds f(x). ## Maxima and Minima Which x makes f(x) as large as possible? Where is the smallest f(x)? - Without calculus, we are reduced to computing values of f(x) and comparing. - With calculus, the information is in df/dx. Suppose the maximum or minimum is at a particular point x. It is possible that the graph has a corner-and no derivative. However, if df/dx exists, it must be zero. The tangent line is level. - The parabolas in Figure 3.3 change from decreasing to increasing. - The slope changes from negative to positive. - At this crucial point the slope is zero. ## 3C Local Maximum or Minimum Suppose the maximum or minimum occurs at a point x inside an interval where f(x) and df/dx are defined. Then f’(x) = 0. **Explanation:** The word "local" allows the possibility that in other intervals, f(x) goes higher or lower. We only look near x, and we use the definition of df/dx. * Start with f(x + Δx) - f(x). If f(x) is the maximum, this difference is negative or zero. The step Δx can be forward or backward: - If Δx > 0: - f(x + Δx) - f(x) is negative - Ax / (f(x + Δx) - f(x)) ≤ 0 and in the limit, df/dx ≤ 0. - If Δx < 0: - f(x + Δx) - f(x) is negative - Ax / (f(x + Δx) - f(x)) ≥ 0 and in the limit, df/dx ≥ 0. Both arguments apply. Both conclusions df/dx < 0 and df/dx > 0 are correct. Thus df/dx = 0. Maybe Richard Feynman said it best. He showed his friends a plastic curve that was made in a special way: "no matter how you turn it, the tangent at the lowest point is horizontal." They checked it out. It was true. **Example 3 (continued):** Look back at Figure 3.3b. The points that stand out are not the "ups" or "downs" but the "turns." Those are stationary points, where df/dx = 0. We see two maxima and two minima. None of them are absolute maxima or minima, because f(x) starts at − ∞ and ends at + ∞. ### Example 4: The Function f(x) = 4x³ - 3x⁴ This function has a slope of 12x² - 12x³. That derivative is zero when x² equals x³, at the two points x = 0 and x = 1. To decide between minimum and maximum (local or absolute), the first step is to evaluate f(x) at these stationary points. - We find f(0) = 0 and f(1) = 1. Now look at large x. The function goes down to − ∞ in both directions. (You can mentally substitute x = 1000 and x = −1000). For large x, −3x⁴ dominates 4x³. **Conclusion:** f = 1 is an absolute maximum. f = 0 is not a maximum or minimum (local or absolute). We have to recognize this exceptional possibility, that a curve (or a car) can pause for an instant (f' = 0) and continue in the same direction. The reason is the “double zero” in 12x² – 12x³, from its double factor x². **Figure 3.4:** The graphs of 4x³-3x and x + x⁻¹. Check rough points and endpoints. **Description:** - The figure shows two graphs. The top one is the graph of f(x) = 4x³−3x, and the bottom one is the graph of f(x) = x + x⁻¹. - The graph of f(x) = 4x³−3x has a local maximum near x = 0.75 and a local minimum near x = 1.25. It also has an absolute maximum at, and an absolute minimum as x approaches − ∞. - The graph of f(x) = x + x⁻¹ has a local maximum at x = -2 and a local minimum at x = 1. There is also an absolute minimum at x = 1. The absolute maximum approaches infinite values as x approaches + ∞. ### Example 5: The Function f(x) = x + x⁻¹ for x > 0 This function has a derivative of 1 - 1/x², which equals zero at x = 1. f(1) = 2 is the minimum value. Every combination like +3 or 1/3 is larger than min = 2. **Important:** The maximum always occurs at a stationary point (where df/dx = 0) or a rough point (no derivative) or an endpoint of the domain. These are the three types of critical points. All maxima and minima occur at critical points! At every other point df/dx > 0 or df/dx < 0. **Procedure:** 1. Solve df/dx = 0 to find the stationary points f(x). 2. Compute f(x) at every critical point—stationary point, rough point, endpoint. 3. Take the maximum and minimum of those critical values of f(x). ### Example 6: The Function f(x) = |x|: - The minimum is zero at a rough point. - The maximum is at an endpoint. - There are no stationary points. The derivative of y = |x| is never zero. Figure 3.4 shows the maximum and minimum on the interval [−3, 2]. This is typical of piecewise linear functions. **Question:** Could the minimum be zero, if the function never reaches f(x) = 0? **Answer:** Yes, f(x) = 1/(1 + x)² approaches but never reaches zero as x→∞. ### Remark 1: x→ +∞ and f(x) → +∞ These can be avoided when f is continuous on a closed interval a ≤ x ≤ b. Then f(x) reaches its maximum and its minimum (Extreme Value Theorem). But x→ ∞ and f(x)→∞ are too important to rule out. You test x → ∞ by considering large x. You recognize f(x) → ∞ by going above every finite value. ### Remark 2: Note the difference between critical points and critical values. - **Critical points** (specified by x) - **Critical values** (specified by f(x)) For example, the function x + 1/x had the minimum point x = 1 and the minimum value f(1) = 2.