6 pts 4. a. Use the Mean Value Theorem to show that for 0 h Use part (a). abovc, to conclude that for 0 8 pts 8. Sketch the graph of a function that satisfies all of the conditions listed below. f(—x)...

Sol: (4) (a) Let 0 .x y 
 
   
     
Then, ( ) is continuous on [ , ] and differentiable on ( , ).
Applying the Mean Value Theorem yields
1
for some in ( , ).
2
1 1 1
However, , since ( ) is a decreas
2 2 2
f x x x y x y
y x
c x y
y x x
g t
c x t




 
 
   
 
 
 
 
 
 
 
 
 
ing function.
Therefore,
1
< ,
2
<
2
(b) From part (a):
<
2
2 < = , by the difference of two squares.
0 <
Square
y x
y x x
y x
y x
x
y x
y x
x
y x
x y x
y x
y x







 

 
both sides:
0 2
2


2
y yx x
yx y x
y x
yx
  
  

 

Sol: (8)
( ) ( ) means the function is "odd"
Visually, it means that the left side of the graph is a mirror image of the
right side, but then flipped upside down.
Next (0) 0
This already has to be the case
f x f x
f
  

 
 
2
, in order to satisfy the first condition (odd
function), so we don't have any new info from this.
Next: lim
This also tells us (because of the odd function condition) that
Next: lim 0
Our
x
x
f x
f x


 

graph has vertical asymptotes:
The above graph satisfy the all the given conditions.
Sol: (9)

2
2
2 8
16
x
y
x



Sol: (10)
To use the closed interval method, we need to find the values of the function
at the endpoints and also where the derivative of the function equals zero.
The largest y-value of these is the absolute maximum in the interval and the
lowest y-value is the absolute minimum in the interval.

The values of the endpoints are:
2
2sin 0.63
4 4 4 4 2
2sin 2 0.43
2 2 2 2
To get the derivative, we use the power rule and the derivative of
f
f
   
 ...