The derivative of the function f(x) = 2x³ + 3x² + 5x is equal to 6x² + 6x + 5. f'(2) = 6*(2)² + 6*2 + 5 = 24+12+5=41 and f(1) = 2*(1)³ + 3*(1)² + 5*1 = 2+3+5=10. the sum of these two values 41+10= 51. The answer is C.

So, the slope of the tangent line to the graph is zero at (-3/2, -37/4). The answer is B.

(x² – 1) / (x² - 1) (x² + 1) = 1 / (x² + 1) ; 1 / (1 + 1) = 0.5 ; x ≠ 1 . pg. 108 . Correct answer is A.

The equation of tangent line is given by the formula:

y – f(x₀) = f'(x₀)(x – x₀)

x₀=2 and f(x₀)=1 are given.

Since f(x)=x²-2x+1 then f'(x)=2x-2 and f'(x₀)=2*2-2=2 for x₀=2.

Thus the equation for the tangent line will be:

y-1=2(x-2)

y-1=2x-4

y=2x-3

(x – 2) (x + 1) / (x – 2) = x + 1 ; x ≠ 2 . pg. 108. Correct answer is E.

A function is increasing where its first derivative is positive. Thus, we have to find the intervals where the first derivative of function is positive.

f(x)=x³-12x

f'(x)=3x²-12

So the function is increasing where f'(x)>0

Namely:

3x²-12>0

3x²>12

x²>4

This is possible when x>2 and when x<-2

Thus the function is increasing in the interval (-∞, -2)U(2,∞)

We know that the total revenue function is given by R=p*x,

Therefore, the answer is E.

(x² + x k x + k² x²) / (x + x k x + k x) = (x² (1 + k + k²)) / (x (k x + 1 + k)) = (x (1 + k + k²)) / (k x + 1 + k) = 0. pg. 182. Correct answer is B

The break-even point is the point at which cost or expenses and revenue are equal: there is no net loss or gain, and one is said to have “broken even.”

f(x)=x²-4x+12

f'(x)=2x-4

f'(x)=2x-4=0 ›x=2

f''(x)=2, 2>0 local minimum.

f(2)=4-8+12=8

at (1, -2, -1) : ?F / ?x = z = -1 ; ?F / ?y = -y = 2 ; ?F / ?z = x – y = 3 ; -1 (x – 1) + 2 (y - (-2)) + 3 (z - (-1)) = -(x – 1) + 2 (y + 2) + 3 (z + 1) = -x + 1 + 2 y + 4 + 3 z + 3 = -x + 2 y + 3 z + 8 = 0 . pg. 187. Correct answer is A.

Break-even point is the point where the total cost and the total revenue are equal C(x)=R(x).

For our question, R(x)=p*x=3x and C(x)=v*x+a=1,5x+2750. 

1,5x+2750=3x,   1,5x=2750,   x=1833,33. The answer is B.

The first derivative must be equal to zero and the second derivative must be positive for a minimum value of a function. Thus:

f(x)=x²-6x+5

f'(x)=2x-6=0 so 2x=6, x=3 and when x=3 y=f(x)=9-18+5=-4

The second derivative is:

f''(x)=2>0

Thus for given function, point (3,-4) is the minimum.

Since the profit is revenue minus the cost we have P(x)=360x-(4800+120x)=240x-4800. Now, to find whether the company loses or makes money insert x=30 into the profit function: P(30)=240·30-4800=+2400, which means that the company gains 2400 TL.

we need to factorise the ratio which follows as;

f(x)=(x²-1)/(x-1)=(x-1)(x+1)/(x-1)=x+1

and if we put "1" instead of x, then the answer will be 1+1=2.

by fractionating the function f(x)=(x-2)(x+2)/(x-2)=x+2 then lim x›2 x+2= 4