1. Continuity and Differentiation
Chapter 2

2. Notes on Notation
𝑓 𝑥 → a continuous function 𝑓 → standard differentiation and continuity. 𝑦 → a possible non-functional expression → requires implicit differentiation

3. Basic Rules of Differentiation
Constant Rule: 𝑑 𝑑𝑥 𝑐 =0 Power Rule: 𝑑 𝑑𝑥 𝑥 𝑛 =𝑛 𝑥 𝑛−1 Product Rule: 𝑑 𝑑𝑥 𝑓 𝑥 ∙𝑔 𝑥 =𝑓 𝑥 𝑔 ′ 𝑥 +𝑔 𝑥 𝑓 ′ (𝑥)

4. Quotient Rule: 𝑑 𝑑𝑥 𝑓 𝑥 𝑔 𝑥 = 𝑔 𝑥 𝑓 ′ 𝑥 −𝑓 𝑥 𝑔 ′ 𝑥 𝑔 𝑥 2 Chain Rule: 𝑑 𝑑𝑥 𝑓 𝑥 𝑛 =𝑛 𝑓 𝑥 𝑛−1 ∙ 𝑓 ′ 𝑥

5. Trigonometric Derivatives
𝑓 𝑥 = sin 𝛼 → 𝑓 ′ 𝑥 = cos 𝛼 ∙ 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝑓 𝑥 = cos 𝛼 → 𝑓 ′ 𝑥 =− sin 𝛼 ∙ 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝑓 𝑥 = tan 𝛼 → 𝑓 ′ 𝑥 = sec 2 𝛼 ∙ 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝑓 𝑥 = csc 𝛼 → 𝑓 ′ 𝑥 = −csc 𝛼 cot 𝛼 ∙ 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝑓 𝑥 = sec 𝛼 → 𝑓 ′ 𝑥 = sec 𝛼 tan 𝛼 ∙ 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒 𝑓 𝑥 = cot 𝛼 → 𝑓 ′ 𝑥 =− cot 2 𝛼 ∙ 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑎𝑛𝑔𝑙𝑒

6. Intermediate Value Theorem (IVT) If 𝑓 𝑥 is continuous on a closed interval [𝑎,𝑏] and if 𝑓(𝑐) is any number between 𝑓(𝑎) and 𝑓(𝑏), then there is at least one 𝑐 on [𝑎,𝑏] such that 𝑎≤𝑐≤𝑏.
𝑓(𝑥)
𝑓(𝑏) 𝑏
𝑓(𝑐)
F Period Finished HERE!! 𝑐
𝑓(𝑎) 𝑎

7. The IVT is used to prove there is a root or zero (x-axis intercept) between two different values.
If 𝑓 𝑎 >0 and 𝑓 𝑏 <0 or if 𝑓 𝑏 >0 and 𝑓 𝑎 <0
𝑓(𝑥)
𝑓(𝑎) 𝑎
Then, there is at least one 𝑓 𝑐 =0
𝑓(𝑐) 𝑐
𝑓(𝑏) 𝑏

8. EX1: Determine if the function has a zero on the given interval
EX1: Determine if the function has a zero on the given interval. 𝑓 𝑥 =− 𝑥 2 −3𝑥+2, [−5,0] 𝑓 −5 =− −5 2 −3 −5 +2 𝑓 −5 =− 𝑓 −5 =−8 𝑓 0 =− 0 2 − 𝑓 0 =2 Since 𝑓 𝑎 <0 𝑎𝑛𝑑 𝑓 𝑏 >0 then there must be a zero on the interval [−5,0]

9. The function 𝑓 is continuous and differentiable on the closed interval [1,5]. The table below gives selected values of 𝑓 on this interval. Which of the following statements must be TRUE? 𝑥 1 2 3 4 5
𝑓(𝑥) -2
(a) 𝑓 ′ 𝑥 >0 𝑓𝑜𝑟 1<𝑥<3
(b) 𝑓 ′′ 𝑥 <0 𝑓𝑜𝑟 3<𝑥<5
(c) The maximum value of 𝑓 on 1,5 must be 5.
(d) The minimum value of 𝑓 on 1,5 must be −2.
(e) There exists a number 𝑐, 1<𝑐<5 for which 𝑓 𝑥 =0

10. With the exception of the end points of a closed interval it, it can be said that: 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑖𝑙𝑖𝑡𝑦 →𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 EX1: If 𝑓is continuous and differentiable find a possible value of 𝑎 and 𝑏 for: 𝑓 𝑥 = 3−𝑥, 𝑥<1 &𝑎 𝑥 2 +𝑏𝑥, 𝑥≥1

11. 𝑓 𝑥 = 3−𝑥, 𝑥<1 &𝑎 𝑥 2 +𝑏𝑥, 𝑥≥1 Continuity → at 𝑥=1, 3−𝑥=𝑎 𝑥 2 +𝑏𝑥 3− 1 =𝑎 1 2 +𝑏(1) 2=𝑎+𝑏 Differentiable → at 𝑥=1, the slope from the left and the right approach the same number. 𝑐𝑎𝑙𝑙 ℎ 𝑥 =3−𝑥 𝑎𝑛𝑑 𝑔(𝑥)=𝑎 𝑥 2 +𝑏𝑥 𝑎𝑡 𝑥=1, ℎ ′ 𝑥 = 𝑔 ′ 𝑥

12. 𝑓 𝑥 = 3−𝑥, 𝑥<1 &𝑎 𝑥 2 +𝑏𝑥, 𝑥≥1 𝑎𝑡 𝑥=1, ℎ ′ 𝑥 = 𝑔 ′ 𝑥 −1=2𝑎𝑥+𝑏 −1=2𝑎(1)+𝑏 𝑠𝑖𝑛𝑐𝑒 2=𝑎+𝑏→𝑏=2−𝑎 −1=2𝑎+ 2−𝑎 𝑎=−3 𝑏=5

13. Practice: 1.) For what values of 𝑚 is 𝑓 𝑥 = sin 2𝑥 , 𝑥<0 &𝑚𝑥, 𝑥≥0 (a) Continuous at 𝑥=0? (b) Differentiable at 𝑥=0? 2.) Graph the function: 𝑓 𝑥 = 𝑥, 0≤𝑥<1 &2−𝑥, 1<𝑥≤2 (a) is 𝑓 continuous at 𝑥=1? Explain. (a) is 𝑓 differentiable at 𝑥=1? Explain.