Lesson 4-QR2

1. Lesson 4-QR2 Quiz 2 Review

2. Objectives • Prepare for the quiz on sections 4-5 thru 4-7 • Graphing • Optimization

3. Vocabulary • Critical number – a value of x such that f’(x) = 0 or f’(x) does not exist • Extreme values – maximum or minimum functional values (y-values) • Inflection point – a point (x,y) on the curve where the concavity changes

4. Closed Interval Method: To find the absolute maximum and minimum values of a continuous function f on a closed interval [a,b]: Find the values of f at the critical numbers of f in (a,b) (the open interval) Find the values of f at the endpoints of the interval, f(a) and f(b) The largest value from steps 1 and 2 is the absolute maximum value; the smallest of theses values is the absolute minimum value.

5. Review of 1st and 2nd Derivatives • Function • Extrema are y-values of the function! • First Derivative – f’(x) • Slope of the function • f’(x) = 0 at “critical” values of x • Possible locations of relative extrema • Relative extrema can also occur at endpoints on closed intervals [a,b] • Second Derivative – f’’(x) • Concavity of the function • f’’(x) = 0 at possible points of inflection • IPs are places where there is a change in concavity

6. Relative Min Relative Max f’ > 0 for x < c f’ < 0 for x < c f’ < 0 for x > c f’ > 0 for x > c x = c x = c 1st and 2nd Derivative Tests • First derivative test uses the change in signs of the slopes [f’(x)] just before and after a critical value to determine if it is a relative min or max • Second derivative test uses a functions concavity at the critical value to determine if it is a relative min or max Slope - 0 + valley Slope + 0 - hill Relative Max Relative Min x = c x = c f’’(x) < 0 Concave down f’’(x) > 0 Concave up

7. Intervals - Table Notation f(x) = x4 – 4x3 = x3(x – 4) f’(x) = 4x3 – 12x2 = 4x2(x – 3) f’’(x) = 12x2 – 24x = 12x (x – 2) X values are listed as in a number line representing the domain. X= values are x-intercepts, vertical asymptotes, critical numbers and possible IPs

8. Graphing Checklist Calculator Helps Here • Domain – for which values is f(x) defined? • x -intercepts – where is f(x) = 0? • y -intercepts – what is f(0)? • Symmetry • y-axis – is f(-x) = f(x)? • Origin – is f(-x) = -f(x)? • Period – is there a number p such that f(x + p) = f(x)? • Asymptotes • Horizontal – does or exist? • Vertical – for what is ? for what is ? Division by 0 or negatives under even roots Type in solve(f(x)=0,x) Type in f(x) | x = 0 Even functions Odd functions Trig functions Limit as x→±∞ F3, limits Type in Lim(f(x),x,a) Division by 0 (and not removed by canceling)

9. Graphing Checklist (cont) • Derivative Information: • Critical numbers – where does f’(x) = 0 or DNE? • Increasing – on what intervals is f’(x) ≥ 0? • Decreasing – on what intervals is f’(x) ≤ 0? • Local extrema – what are the local max/min? Use f’ or f’’ test. • Concavity • Up – where is f’’(x) > 0? • Down – where is f’’(x) < 0? • Inflection points – where does f change concavity? F3 dif(f(x),x) Copy derivative and paste into solve(f’(x)=0,x) Type in f’(x) | x = value 2nd DT: Type in f’’(x) | x = critical # F3 dif(f’(x),x) Copy derivative and paste into solve(f’’(x)=0,x) Type in f’’(x) | x = value Use calculator to check your info by graphing the function. Be Careful: the small screen can lead to some tricky views

10. Example 1 f’(x) = -2x/(x² - 4)² 1 Graph -------------- x² – 4 f’’(x) = 2(3x² + 4)/(x² - 4)³ Domain: x –intercepts: y –intercepts: Symmetry: Y-axis: Origin: Periodic: Asymptotes H: V: Critical numbers: Increasing: Decreasing: Max/Min: Concavity x ≠ ± 2 None, y ≠ 0 y = -1/4 Yes No No x = -2, 2 y = 0 x = 0 x < 0 x > 0 At x = 0, y = -1/4 is a relative max Up: |x|>2 Down: |x| < 2

11. Example 1 Graph y x Yellow lines are asymptotes. Blue curve represents the curve you should have drawn Pink curve represents the curve your calculator should have drawn. The small screen size and your window limits can cause issues!

12. Procedures for Solving Optimization Problems • Make a sketch. Label the picture with the quantities given. • Find a primary expression for the quantity to be optimized. • Find a secondary relation from the given in the problem. • Use the secondary relation to reduce primary equation to a single variable equation. • Determine the domain for the single variable, usually a closed interval. • Find the critical points and check the functional values at the end points. • Use the techniques of calculus (1st or 2nd Derivative Tests) to determine the type of extrema critical number represents. • Answer the question (including units).

13. Optimization Example 1 Govt Fencing Farmer’s Field A farmer whose property borders I-81 wants to maximize the size of the pasture for cattle he can create with his limited resources. He can only afford to buy materials for 3000 feet of fencing. How big can he make his pasture? A = l• w and l + 2w = 3000 so l = 3000 – 2w A = (3000 – 2w) • w = 3000w – 2w² dA --- = 3000 – 4w = 0 when 4w = 3000 or w = 750 ft dw A = 750 • (3000 – 1500) = 1,125,000 sq ft

14. Example 2 Find two positive numbers that minimize the sum of three times the first number plus the second number, if the product of the two numbers is 192. Primary Equation: S = 3x + y Secondary Equation: 192 = x ∙ y S = 3x + y S = 3x + 192 / x 192 = x ∙ y 192 / x = y dS/dx = 3 – 192 / x² S’ = 3 – 192 / x² = 0 when 3 = 192 / x² 192 = 3x² s = ± 8 (the minus has no meaning) S’’ = 384 / x³ concave up for all x > 0 so a minimum Numbers: x = 8, y = 192 / x = 24