Chapter 21 Application of Differential Calculus DP Studies Y2 Chapter 21 Application of Differential Calculus

Contents A. Increasing and decreasing functions B. Stationary points C. Rates of change D. Optimization

A: Increasing and Decreasing Functions Quick reminder of notations for inequalities an intervals.

A: Increasing and Decreasing Functions

A: Increasing and Decreasing Functions We can determine intervals where a curve is increasing or decreasing by considering f'(x) on the interval in question. For most functions that we deal with in this course:

A: Increasing and Decreasing Functions Many functions are either increasing or decreasing for all x є R . We say these functions are monotone increasing or monotone decreasing.

A: Increasing and Decreasing Functions for a strictly increasing function, an increase in x produces an increase in y “Monotone increasing is an uphill battle all the way.”

A: Increasing and Decreasing Functions for a strictly decreasing function, an increase in x produces a decrease in y. “It’s all downhill with monotone decreasing.”

Example 1 Find the intervals where f(x) is: Increasing decreasing

A: Increasing and Decreasing Functions a. The increasing part is as x increase, y also increase. b. The decreasing part is as x increase, y decrease.

Sign Diagrams Sign diagrams for the derivative are extremely useful for determining intervals where a function is increasing or decreasing. derivative = slope If the derivative is negative, the function is decreasing. If the derivative is positive, the function is increasing. The critical values for f’(x) are the values of x for which f’(x) = 0 or f’(x) is undefined. When f’(x) = 0, the critical values are shown on a number line using tick marks. When f’(x) is undefined, the critical values are shown with a vertical dotted line.

Sign Diagrams turning point (derivative =0)

Sign Diagrams

Sign Diagrams Example 2: Find the intervals where f(x) = 2x3 + 3x2 – 12x – 5 is increasing or decreasing.

Sign Diagrams Solution to example 2: f(x) = 2x3 + 3x2 – 12x – 5 f(x) = 6x2 – 6x – 12 Set f(x) = 0 and solve by factoring or quadratic formula 6(x2 – x – 2) = 0 6(x – 2)(x + 1) = 0 => x = 2, x = -1

C: Stationary points A stationary point of a function is a point where f’ (x) = 0. It could be a local maximum, local minimum, or stationary inflection.

Example 3 Find and classify all stationary points of f(x) = x3 – 3x2 – 9x + 5

Stationary points at x = 3 and x = -1 Determine the interval of increasing and decreasing by choosing x-values less than -1, between -1 and 3, and greater 3. x < -1: (-2 – 3)(-2 + 1) = (–)(–)=+ -1 < x < 3: (0 – 3)(0 + 1) = (–)(+)=– x > 3: (4 – 3)(4 + 1) = (+)(+) = + + – + increasing -1 decreasing 3 increasing local maximum local minimum

+ – + increasing decreasing increasing -1 3 local maximum local minimum

Example 4: Find the greatest and least value of y = x3 – 6x2 + 5 on the interval -2 < x < 5.

Take the derivative of the function. Find the stationary points. Draw the sign diagram. Determine the local max. and local min. within the given interval.

4 3: Draw the sign diagram + + – increasing decreasing increasing 4 local maximum local minimum

Check values Check y-values at the stationary points and the endpoints. Stationary point, x = 0 y = 03 – 6(0)2 + 5 = 5 Stationary point, x = 4 y = 43 – 6(4)2 + 5 = -27 Left endpoint, x = -2 y = 23 – 6(-2)2 + 5 = -27 Right endpoint, x = 5 y = 53 – 6(5)2 + 5 = -20 The greatest value, 5, occurs at x = 0. The least value, -27, occurs at x = 4 and at x = -2

C. Rate of Change Example 5:

C. Rate of Change Solution to example 5:

C. Rate of Change

C. Rate of Change Example 6

C. Rate of Change

D. Optimization There are many problems for which we need to find the maximum or minimum value of a function. The solution is often referred to as the optimum solution and the process is called optimization.

D. Optimization We can find optimum solutions in several ways: using technology to graph the function and search for the maximum or minimum value using analytical methods such as the formula x =-b/2a for the vertex of a parabola using differential calculus to locate the turning points of a function. These last two methods are useful especially when exact solutions are required.

Warning!! The maximum or minimum value does not always occur when the first derivative is zero. It is essential to also examine the values of the function at the endpoint(s) of the interval under consideration for global maxima and minima.

D. Optimization

D. Optimization Example 7: A 4 liter container must have a square base, vertical sides, and an open top. Find the most economical shape which minimizes the surface area of material needed.

D. Optimization

D. Optimization

D. Optimization

Example 8: A square sheet of metal 12 cm x 12 cm has smaller squares cut from its corners as shown. What sized square should be cut out so that when the sheet is bent into an open box it will hold the maximum amount of liquid?

V’ = 12x2 – 96x + 144 V’ = 12(x2 – 8x + 12) V’ = 12(x – 6)(x – 2) Set the derivative to 0 and solve x – 6 = 0  x = 6 x – 2 = 0  x = 2 If we put the x-values back into the original equation we get V = x(12 – 2x)2 at x = 6, V = 6(12 – 2(6)2) = 0 at x = 2, V = 2(12 – 2(2)2) = 8 Therefore, the only value for x is 2. The square we are cutting out is a 2cm x 2cm.

Example 9:

Solution to example 9: