MATEMATİK I

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