The Chain Rule: The derivative of composite function f (g(x)) is f'(g(x)) * g' (x). For this question our g(x) = 2x+1. The derivative of (2x + 1)³ yields to 3(2x + 1)² * 2 which is equal to 6(2x + 1)² . The answer is D.

Because x approaches -3 from the right side x is more than -3 and for x > -3, f is given by the rule f (x)=–x+12. Hence, the result is as follows. 

lim_x›-3⁺   f(x) = -(-3) + 12 = 3 + 12 = 15

?f / ?x = 2x + 2 = 0 ; x = -2 ; ?f / ?y = 4 y + 3 = 0 ; y = -3 / 4 ; a = ?²f / ?x² = 2 > 0 ; b = ?²f / ?y² = 2 ; c = ?²f / ?x ?y = 0 ; D = a . b – c² = 2 . 2 – 0 = 4 > 0 ; pg. 192. Correct answer is D.

x = C \ B = {2, -2, -3, -5} ; x ? A = {-3, -5}. pg. 4. Correct answer is A.

f'(x)=(-3x+11)'=-3, f'(0)=-3. So, the answer is A.

f(x)=3x.e^-5x

f'(x)=(3x)'.e^-5x+3x.(e^-5x)'

f'(x)=3.e^-5x+3x.(-5e^-5x)

f'(x)=(3-15x)e^-5x

The velocity at time t is the derivative of function f(t) at time t which is defined by the limit given in C.

The break-even point is where the revenue and cost are equal. Here, revenue is R(x)=360x and the cost is C(x)=4800+120x. Equating them we find 360x=4800+120x and therefore the break-even point is x=20.

Let x and y be the side lengths of the rectangele. If the perimeter is 80, then 2x+2y=80

x+y=40

y=40-x

The are of the rectangle is A=x*y=x*(40-x)=40x-x²

Since it is a maximization problem the first derivative of the area function (A) must be equal to zero and the second derivative must be negative.

A''(x)=-2<0 (Second derivative rule satisfied)

A'(x)=40-2x=0, 40=2x, 20=x, y=40-x=20 So the area A=20*20=400

Let x denote the number of items produced and sold. The total revenue is
R (x) = 130 x

and the total cost is
C (x) = 50 x + 4400

where, now, ?=45. To find the break-even point we equate total cost to total revenue, so

C(x)=R(x)? 50 x+4400=130 x?80 x=4400
so that
x =55
Thus, the manufacturer has to sell at least 55 units to break even, i.e. no profit, or no loss.

Since a rational function f is continuous on the domain of its definition, f is continuous on its domain. the  is continuous for all real numbers x except -1 and 1. Thus, f is continuous on             R\{-1, 1}. The answer is B.

The answer is C.

f(2)=7.2+1=15

g(2)=2²-1=4-1=3

f/g(2)=15/3=5

R \ {1} ; (x³ – 1) = 0 ; x = 1. pg. 117. Correct answer is B.

y = 0.25 + 2 = 2.25. pg. 109. Correct answer is C.