-∞ ; Since the degree of the numerator is greater than the degree of the denominator. pg. 114. Correct answer is D.

From the product rule for limits, we have

lim_x›2 (x + 3) (2x - 5) = lim_x›2 (x + 3) . lim_x›2 (2x - 5)

                              = (2 + 3) . (4 - 5)

                              = -5

f (x) = x² -1 is a polynomial function. Hence, it is a continuous function, and its set of continuity is R. 

Let Q(t) be the amount of the radioactive substance at time t and A be the amount of initial radioactive substance. Since decay of the amount of radioactive substance is exponential, the function Q(t) has the following form:

Q(t) = Q₀e^kt

for t=0›A= Q₀   then Q(10)=0,95A=A.e^k10

ln 0,95=10k and k=(ln0,95)/10

To find the half-life of this radioactive material we should solve the equation;

0,5A=A.e^{((ln0,95)/10).t}

ln0.5=(ln0,95/10)).t ›135,13

Since the perimeter is x + 2y and the total fence available is 500 m we write x + 2y = 500. From this, x is found as,

x=500-2y and the area of this rectengular is x.y=(500-2y).y = 500y-2y²

f(y)=500y-2y²

f'(y)=500-4y

f'(y)=500-4y=0 (for local maximum value)

y=125m and x=250 m

max area is x.y=125m*250m=31250 m²

f(x)=x²+7x-8

f'(x)=2x+7

f'(2)=11

Since x approaches 0 from the left side, x is negative and for x<0, |3x|=-3x is true. So, we get following result.

. The answer is C.

The answer is B.

f(x,y)=x⁴+x²y²-2x²+2y²?

We have to find the points simultaneously satisfy fx=0 and fy=0

fx=4x³+2xy²-4x=0, so x(4x²+2y²-4)=0

fy=2x²y+4y=0, so y(2x²+4)=0

Points given in III and V doesnt satisfy these conditions simultaneously. So the remaining 3 points are stationary points.

let x denote the demand and p the price of the product. We are given (x₁,p₁)=2000,1200) and (x2, p2)=(3000,1000), since there is TL discount in the price. The demand line that passes through these points has slope m=-1/5 and its equation is p=-x/5 +1600

(x – 1) / (x – 1) = -1 . pg. 109. Correct answer is D.

f(g(x))=(2x+1)^1/2

From the chain rule we know that (f(g(x)))'=f'(g(x))g'(x)

Applying the chain rule we get (f(g(x)))'=(1/2)(2x+1)^-1/22=(2x+1)^-1/2

Maximum profit is calculated by taking the derivative of the profit function, in this case, P' (x)=16-0,04x. Equating this derivative to zero we get x=400. Since the function is decreasing on the left and right of this point the required maximum point and the value of the profit at this point is P(400)=3200.

f(x)=x²+2x+lnx

f'(x)=2x+2+(1/x) and for x=1 f'(1)=2+2+1=5

From the quotient rule for limits together with  lim_x›1 (2x - 8) ? 0, we obtain

(1² + 5 . 1 - 12) / (2 . 1 - 8) = -6 / -6 = 1.