2. ADDITIONAL PROBLEMS 5 points each: I) Define f: R - R as follows:f ( x ) : ={0 if x 0Show that...

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2. ADDITIONAL PROBLEMS 5 points each: I) Define f: R -> R as follows:
f ( x ) : =
{0 if x < 0 e 1- /x if x > 0
Show that f ' (0) = 0 and f " (0) = 0. You may use without justification the fact that limx--.00 xk /ex = 0 for all real numbers k.' z) Use the chain rule to show that cost x + sin2 x is constant. You may use the fact that the only functions R -+ R with identically zero derivatives are constant functions.2) Conclude, by choosing a value for x, that cos' x + sin2 x = 1 for all x.
'One can show that f (n) (0) = 0 for all nonnegative integers n. Consequently, f is an example of a function all of whose derivatives equal zero at 0, yet f is not identically zero in any open interval containing 0. Hint: It is not valid to compute f' (x) on (—oo, 0) and (0, oo) and then take the limit as x —¦ 0. This does not take into account the value off at x = 0, which is essential if f is even going to be continuous at x = 0. One must use the definition of the derivative to compute f'(0). Then one can compute

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David · Accepted Answer

Question
Solution 
1) 
'
0
( ) ( )
( ) h
f x h f x
f x Lim
h

 
  
'
0
( ) (0)
(0) h
f h f
f Lim
h


  
1/
'
0
0
(0)
h
h
e
f Lim
h



  
'
0 1/
1/
(0) h h
h
f Lim
e
  
Now take y= 1/h 
Then when h tends to zero y tends to infinity 
' (0) y y
y
f Lim
e
  
' (0) 0( )y y
y
f Lim given
e
   
' 1/
2
1
( ) xf x e
x
       for x>0   
'( ) 0f x                    for x0   
'( ) 0f x                    for x

2. ADDITIONAL PROBLEMS 5 points each: I) Define f: R -> R as follows: f ( x ) : = {0 if x 0 Show that f ' (0) = 0 and f " (0) = 0. You may use without justification the fact that limx--.00 xk /ex = 0...

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