df(x)/dx=f '(x)=a*n*x^(n-1)-b=anx^(n-1)-b

Since the denominator of the expression at x = 2 is 0, we must look at the one-sided limits in order to find the limit at x = 2 if it exists. Then we have,

lim_x→2⁺  (  x  / |x-2| ) =  lim_x→2⁺  (  2  / 0⁺ ) = +∞

lim_x→2^-  (  x  / |x-2| ) =  lim_x→2^-  (  2  / 0⁺ ) = +∞

Hence, since the one-sided limits show the same behavior at x = 2, the limit exist at x =2, and it is +∞.  

The profit is the revenue minus the cost. In this case the revenue is the demand multiplied by the price giving R(Q)=(32-2Q)Q=32Q-2Q². Inserting revenue and cost into the equation for profit we obtain P(Q)=32Q-2Q²-2Q²=32Q-4Q².

The answer is A.

A quadratic function is a polynomial function of degree2. Thus the largest exponent must be 2, and all the exponents must be natural numbers. Only the function in E satisfies these conditions.

E_p = |(p / q) (dq / dp)| = ; dq / dp = -0.06 p ; E_p = |(p / (492 – (0.03 p²))) (-0.06 p)| = 0.06 p² / (492 – (0.03 p²)) ; p = 20 ; E_p = 0.06 x 400 / (492 – 0.03 x 400) = 24 / (492 – 12) = 24 / 480 = 0.05 . pg. 160. Correct answer is A.

We are asked to find the derivative of the total revenue function at the point q=10. So, T'(p)=400·2q-2·3q²=800q-6q². Then T'(10) = 800·10-6·10²=7400.

in the simplest terms the definition of demand tells us the desire or willingness of a customer to pay a certain price for a particular item she wants to buy. Suppose that the seasonal price of a pair of sneakers is 180 TL. The same sneakers cost 100 TL in the discount season. A customer therefore may prefer to buy it at the discount season, and even maybe buys two pairs instead of one pair. Therefore, demand is proportional to the price of the goods.

2 ; Since the degree of the numerator is equal to the degree of the denominator. pg. 114. Correct answer is C.

The derivative of  f(x) = 2x³ + 5x + 3 is equal to f'(x) = 6x² + 5. At point x=3, f'(3) = 6*(3)² + 5 = 54 + 5 = 59. The answer is C.

Second derivative test for local extrema: if a is a critical point and the second derivative f ¨(a) is
positive, i.e. f ¨(a) > 0, then a is a local minimum point. If f ¨(a) is negative, i.e. f ¨(a) < 0, then a
is a local maximum point. For f(x) the derivative is equal to f'(x)=4x+2 and the second derivative is equal to f''(x)=4 this value is a positive value which indicates a local minimum point. If we equal the derivative to zero, f'(x)=4x+2=0 the x value would be equal to -1/2. The answer is E.

The answer is B.

The point where the quantity demanded equals the supply provided is called the equilibrium point.

Since 1/2, 1/3 and -1 are not natural numbers, A, C, D and E are not polynomial functions. All powers of the variable x must be natural numbers. The only available function that satisfies this condition here is f(x) = 2^1/2 x +1, where the power of x is 1.      

The answer is E.

-3 x + 3.333 = 5 x + 333 ; 8 x = 3.000 ; for surplus : x > 250 . pg. 160 . Correct answer is E.

Let one side be x and other be y. Then 2x+2y=40 if the perimeter is 40. We want to find the maximum of f(x,y)=xy given perimeter is 40. We can decrease the number of unknowns by substitution using the perimeter equation.

2y+2x=40, y=20-x

f(x,y)=g(x)=x(20-x)=20x-x² will be the function of one variable to be maximized.

g_x=20-2x=0 then x=10, since y=20-x=10. So the area will be 10*10=100

Because x approaches 1 from the left side x is less than 1 and for x<1, f is given by the rule f(x)=-2x+5. So, the result follows as

. The answer is E.