The function is a rational function. Since the degree of the numerator is less than the degree of the denominator, the following result is true:

. The answer is E.

 The answer is D.

Once again price elasticity of demand is

The answer is A.

The answer is A.

Let us write the demand function for q

q=2700-3p

Taking the derivative with respect to p we get

dq/dp = -3

The price elasticity of demand is

From this equation we can easily write the price for the unit elasticity which is 

1=3p/(2700-3p)

and the sought for price is p=450. Therefore, the answer is C.

(f ° g)(2) equals to f ( g(2) ). g(2), substituting 2 with x in the g function is -2 +3 = 1. Then, (f ° g)(2) = f (1) = 3 . 1 -7 = 3 - 7 = -4. 

Price elasticity of demans is 

xy=6050

L(x, y, ?) = x+2y+ ?(xy-6050).

?L/?x =0, ?L/?y =0, ?L/?? =0

1+?y=0, 2+?x=0, xy-6050=0

Hence, ?=-1/y=-2/x

x=2y

2y²=6050

y=55, x=110

x+2y=220 m.

The additional cost needed to produce or purchase one more unit of a good or service is called the marginal cost. It corresponds to the derivative of the total cost function C(x).

f(x) = (x+1)(3-x)

f'(x) = (x+1)'(3-x)+ (x+1)(3-x)'

f'(x) = (3-x)+(-x-1)

f'(x) = 2-2x

f'(x) =2-2x=0 › x=1

f(1) = 4

It is clear that this function has an inverse. Hence, for finding f^-1 (x), we write y = 3x + 5. What we find is x = (5 - y) / 3. Then, f^-1 (x) = (5 - y) / 3. Substituting 2 with x, we find 1.  

E_p = |(p / q) (dq / dp)| = 1 for unit elastic ; q = 2.575 – (5 p) ; dq / dp = -5 ; E_p = |(p / (2.575 – (5 p))) (-5)| = 1 ; p / (2.575 – (5 p)) (-5) = 1 ; 5 p / (2.575 – 5 p) = 1 ; 10 p = 2.575 ; p = 257.5 . pg. 166. Correct answer is C.

If the company wants to make a profit of 2500TL then the profit function must be equal to this quantity. In other words,

P(x)= 2500 ? 80 x-4400=2500?80 x=6900 so that
x=86,25 units.

f(x,y)=(3lnx+1)/(e^-3x) + (y²x²)/(x+y)

?f / ?y = [(2x²y).(x+y)-(x+y).1] / (x+y)²

By the way x=1,y=-1

?f / ?y = [(4).(3)-(3).1] / (3)²=1

Becuase x approaches 3 from the right side x is greater than 3 and for x>3, f is given by the rule . So, the result follows as

. The answer is A.

If x² -6x + 5 = 0 then the roots are {1, 5} from the formula (see p. 54). Since the left side of 1 and the right side of 5 on the sign table are positive (+), the solution of x² -6x + 5 < 0 is the interval (1, 5).

e^y + x² . pg. 183. Correct answer is D.

The profit function we would like to maximise is P(x)=R(x)-C(x)
R(x)=x p(x)=x ·(200-3x)=200x -3x²

so that

P(x)=-2x ² +120x-75
To find the maximum price, let us take the derivative of P(x) and equate it to zero:

P'(x)=-4x+120 = 0 ? x=30

(x – 1) (x² + x + 1) / (x – 1) = x² + x + 1 = 3 ; x ≠ 1 . pg. 108. Correct answer is D.