MATH 1401 #10 QUESTION 3

#10: 3a) ANSWER: (-1,0), (0,1) and (1,0)

SOLUTION: dy/dx = D _x ( ( 1 - x ² ) ^½ ) = (½)( 1 - x ² ) ^{- ½} (-2x) = x ( 1 - x ² ) ^{- ½} .

dy/dx = 0 when x = 0 and y = 1. dy/dx is undefined at 1 - x ² = 0 or x = + 1 and y = 0.

#10: 3b) HOMEWORK #5

#10: 3c) ANSWER: (/2 + 2n,1) and (- /2 + 2n,-1)

SOLUTION: dy/dx = D _x (sin x) = cos x = 0 implies x = + /2 + 2n.
y = 1 at x = + /2 + 2n and y = -1 at x = - /2 + 2n.

#10: 3d) HOMEWORK #6

#10: 3e) ANSWER: ((1/3) ^½ , - (2/3)(1/3) ^½ ), ( - (1/3) ^½ , (2/3)(1/3) ^½ ), (-1,0) and (3,24)

SOLUTION: dy/dx = D _x ( x ³ - x ) = 3x ² - 1 = 0 implies x = + (1/3) ^½ which are both in the given interval. At x = (1/3) ^½, y = - (2/3)(1/3) ^½ ; and at x = - (1/3) ^½ , y = (2/3)(1/3) ^½. The derivative is not defined at the endpoints so x = -1 and 3 also correspond to critical points. At these values y = 0 and 24, respectively.

#10: 3f) HOMEWORK #7

#10: 3g) ANSWER: (0,0)

SOLUTION: so dy/dx = . Therefore, the only critical point is when the derivative is undefined at x = 0, y = 0.