Sketch the following sets in R^2 and hence determine whether they are bounded or

Question

Sketch the following sets in R^2 and hence determine whether they are bounded or unbounded, and whether they are convex or not. In the case where the set is not convex, give an example of a line segment where the definition of convexity breaks down. (a) {(x, y) | x + y greaterthanorequalto 1, x - y lessthanorequalto 2, x greaterthanorequalto 0} (b) {(x, y)|1 lessthanorequalto |x| lessthanorequalto 2, |y - 3|lessthanorequalto 2} (c) {(x_1, x_2)|x^2_1 + x^2_2 lessthanorequalto 4 and x_1 + x_2 lessthanorequalto 1} (d) {(x_1, x_2)|X^2_1 + x^2_2 greaterthanorequalto 1 and x_1 + x_2 lessthanorequalto 4 Differentiate the following functions: (a) sin (Squareroot x cos (x)) (b) (2x - 3)^6/x^2 + 1 What is the area under the curve y = |sin (2x) | between x = 0 and x = pi?

Explanation / Answer

2 . a.)

y = sin(underoot(x)cosx)

dy/dx = cos(underoot(x)cosx) [ (1/2*underoot(x) )cosx - sinx *underoot(x) ]

= cos(underoot(x)cosx) [ (1/2*underoot(x) )cosx - sinx *underoot(x) ]

b.)

y= (2x -3)^6 / (x*x + 1 )

dy/dx = [(x*x + 1 ) d/dx(2x -3)^6 - (2x-3)^6 d/dx(x*x +1)] / (x*x + 1)^2

dy/dx =[ (x*x + 1) * 12(2x-3)^5 - (2x-3)^6* 2x ]/ (x*x + 1)^2

3 . y = Integration[ x=0->] |sin(2x)| dx

let 2x = t

2dx = dt

dx = dt/2

y = Integration[t=0->2] |sint| dt/2

y = 1/2[ Integration [t=0->2] |sint| dt]

  

  

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.