MATEMATİK I - Deneme Sınavı - 8

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,∞)

Let's find it by the help of limit:

So the answer is A.

So for given function average velocity in [4,6]=(f(6)-f(4))/(6-4)=((6²+4*6)-(4²+4*4))/(6-4)=28/2=14

f '(t)=4t-4 

Thus for t=3 f '(t)=4t-4=4*3-4=8

f(x)=2x^-1+3x²

f '(x)=2*(-1)x^-1-1+3*2x=-2x^-2+6x=6x-(2/x²)

let f(x)=h(x)g(x) where g(x)=e^x and h(x)=x²

f '(x)=h'(x)g(x)+h(x)g'(x)

Since h'(x)=2x and g'(x)=e^x

h'(x)=2xe^x+x²e^x

f '(x)=6x²-(1/x²)

f ''(x)=12x+(2/x³)

f '(x)=2x³-8x=2x(x²-4)

f '(x)=0 for x=-2, x=2, and x=0

We can rewrite this equation as y=100-x

The area of rectangle is A=xy=x(100-x)=100x-x²

A is maximum when dA/dx=A'(x)=0 and A''(x)<0.

Thus A'(x)=100-2x=0 so x=50 and y=100-x=50. 

So the area A=50*50=2500

Note that A''(x)=-2<0. Thus it is a maximum point.

Thus for f(x)=x²-4x the first derivative is f '(x)=2x-4

Since 0<2x-4

4<2x

2<x

25-1/4p=2p+20

which is written as

25-20=2p+1/4p

and the price is found as

p=20/9. Therefore, the answer is D.

q=1750-5p

Taking the derivative with respect to p we get

dq/dp = -5

The price elasticity of demand is

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

1=5p/(1750-5p)

and the sought for price is p=175. Therefore, the answer is A.

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.

-4x+150=6x-275, x=42,5. Because there will be a shortage in the market, to the left of this point the correct is x<42,5. Therefore, the answer is B.

Therefore, the answer is E.