g(x)=2 sin(x) +e^(x/pi) on [0,2pi]
A= 1/(2pi -0) \(\int_{0}^{2pi}\)(2 sin(x) + e^(x/pi)) dx
= 1/2pi(-2 cos(x) + pi e^(x/pi))
Then plug in 0 and 2 pi in for X
2-pi+(1/2)pi(e^2pi -2)
I know how to set it up, just not 100% sure thats the right answer or not.
Thank you!
Ok good that's what I got haha
Have you done the one where it is asking to find the max area of a right triangle?
V=x^2y = 8 => 8/x^2
C= 1*x^2+4(4xy)
C= x^2+16xy
C(x)= x^2+16x* 8/x^2
C(x)= x^2+128/x
C'(x)= 2x-128/x = 0
2x = 128/x^2
2x^3=128
x^3=64
x=4
C''(x)= 2+256/x^3 >0, for all x>0
By the second derivative test for absolute min, cost is minimum when x=4, y= 1/2
Intervals increasing : (-infinity, infinity)
Intervals decreasing: none
local max : none
local min : none
no local max/min because f(x) is always increasing
intervals concave up : (-infinity, 0)
intervals concave down: (0, infinity)
Thank you very much
4pi/3
5pi/3
1/2
-1/2
You didn't ask, but the function would have a local minimum at A and a local maximum at B.
F'(X)=6x^2+6x-72
6x^2+6x-72=0
x^2+X-12=0
(X+4)(X-3)=0
x=-4 X=3
Then in a chart with X values on one side and f(X) values on the other side (I don't know how to make a chart on here if you can)
0, 3
5, -32
3, -132
So,
absolute maximum of f(X)=3 at X=0
absolute minimum of f(X)= -132 at X=3
R(500)=5,000,000
R'(500)=40,000
R(502)=?
By linear approximation,
L(p)=R(500)+R'(500)(p-500)
L(502)=5,000,000+40,000(2)
= 5,080,000=R(502)
+95 