Solution for WA1-V2KM - LLorem epsum PDF

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Written Assignment Unit 1 Solutions

1. Answer: Use the two-point formula for a linear function, 𝑦 = We have 𝑦 =

𝑓(7)−𝑓(0)

That gives 𝑦 =

7−0 14−8 7

𝑓(𝑥2 )−𝑓(𝑥1 ) (𝑥2 )−(𝑥1)

(𝑥 − 𝑥1 ) + 𝑓(𝑥1 )

(𝑥 − 0) + 𝑓(0) 6

(𝑥 − 0) + 8 = 𝑥 + 8 7

Therefore, the linear function, 𝑓(𝑥) =

6 𝑥+ 7

2. Answer: Using the difference quotient, 2𝑡 − ℎ

8.

𝑓(𝑡+ℎ )−𝑓(𝑡 ) ℎ

=

2(𝑡+ℎ )−(𝑡+ℎ)2+3−2𝑡+𝑡 2−3 ℎ

=

2ℎ−2𝑡ℎ−ℎ 2 ℎ

= 2−

3. Answer: We would like to write the equation in terms of a single trigonometric function. Use Pythagorean identity, 𝑠𝑖𝑛2 𝑥 = 1– 𝑐𝑜𝑠2 𝑥, to reduce the equation as, −4 cos 𝑥 = 𝑐𝑜𝑠 2 𝑥 0 = 𝑐𝑜𝑠 2 𝑥 + 4 cos 𝑥 0 = cos 𝑥 (cos 𝑥 + 4) cos 𝑥 = 0 𝑜𝑟, cos 𝑥 + 4 = 0 For cos 𝑥 = 0, we get solutions at 𝜋/2 and 3𝜋/2 in the interval [0, 2𝜋). 𝜋 This means, all solutions will look like 𝑥 = + 𝑘𝜋, where k is an integer. 2 For cos 𝑥 = −4, there is no solution.

4. Answer: Domain is (−∞, ∞), range is (0, ∞) and horizontal asymptote is 𝑦 = 0.

5. Answer: Domain is (−2, ∞), range is (−∞, ∞) and vertical asymptote is 𝑥 = – 2.

6. Answer: {𝑥|𝑥 ≠– 9, 9}, since denominator can’t be zero. 7. Answer: 𝑓(– 15) = 0 and 𝑥 = – 15, 17.5, 25 8. Answer: Function, and domain: {𝑥| − 𝜋 ≤ 𝑥 ≤ 𝜋} range: {𝑦| − 1 ≤ 𝑦 ≤ 1}, intercepts are 𝜋 𝜋 at the points (−𝜋, 0), (− 2 , 0), (0, 0), ( 2 , 0), (𝜋, 0) 9. Answers: a. odd; b. odd; c. neither 10. Answers: a. $27.50; b. $32.50; c....