TOPICS ON CHAPTER 4 TEST: 1 TOPICS ON CHAPTER 4 TEST: 1. Finding extreme values (Remember, if it’s on a closed interval, you must test the critical points AND the endpoints!) Mean Value Theorem First and Second Derivative Test (and using it to graph functions) Critical Values and Points of Inflection Optimization Newton’s Method Related Rates

SOLUTION: 𝑑𝑏 𝑑𝑡 =3 𝑑ℎ 𝑑𝑡 =−3 𝐴= 1 2 𝑏ℎ 𝑑𝐴 𝑑𝑡 = 1 2 𝑏 𝑑ℎ 𝑑𝑡 +ℎ 𝑑𝑏 𝑑𝑡 𝑑𝐴 𝑑𝑡 = 1 2 𝑏(−3)+ℎ(3) 𝑑𝐴 𝑑𝑡 = −3 2 𝑏+ 3 2 ℎ If b>h, 𝑑𝐴 𝑑𝑡 will be negative!!! This means the area will be decreasing if b>h!

Find the critical values of . 𝑥=4.5, 0, 𝑎𝑛𝑑 9

Use Newton’s method to approximate the x-coordinate of the POSITIVE point of intersection of 𝑦= 𝑥 4 −3𝑥+4 and 𝑦=− 𝑥 2 +8 to FIVE decimal places. SOLUTION: 𝑦= 𝑥 4 −3𝑥+4 and 𝑦=− 𝑥 2 +8 intersect whenever 𝑥 4 −3𝑥+4=− 𝑥 2 +8 or 0= 𝑥 4 −3𝑥+4+ 𝑥 2 −8 This simplifies to 0= 𝑥 4 + 𝑥 2 −3𝑥−4. Starting with a guess of x =1, the table at right shows the values obtained using Newton’s method: Our answer is 1.58073 (since these five decimal places remained constant in the last two calculations!

Find the absolute extrema of SOLUTION:

Sketch the graph of f(x) if….

Find the absolute extrema of on the interval [0,2𝜋]. SOLUTION:

SOLUTION: 𝑓 ′ 𝑥 = 𝑥 2 −8𝑥+12= x−6 x−2 𝑥=6 𝑜𝑟 𝑥=2 Absolute maximum of 22 at x=9. X Y -5 2 5.67 6 9 22

SOLUTION:

SOLUTION: 4 inches

Let f be the function given by 𝑓 𝑥 = 𝑥 3 −3 𝑥 2 Let f be the function given by 𝑓 𝑥 = 𝑥 3 −3 𝑥 2 . What are all values of c that satisfy the conclusion of the Mean Value Theorem on the closed interval [0,3]? SOLUTION: c = 0 and 2

Given a. Find the coordinates of all maximum and minimum points on the given interval. Justify your answers.   b. Find the coordinates of all points of inflection on the given interval, Justify your answers. c. On the axes provided, sketch the graph of the function. Relative max at (-3,0) Relative min at (-1,-4) Inflection point at (-2,-2)