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suppose a function f is differentiable for all x and f(0)=0. If g(x) is defined by g(x)= f(x)cos(x), which of the following statements must be true.

1) there exist a number c in the interval (0,pi/2) such that g'(c)=0

2)there exist a number c in the interval (pi/2,pi) such that g'(c)=0

3)there exist a number c in the interval(-pi/2,0) such that g'(c)=0

Guest Feb 22, 2015

Best Answer 

 #2
avatar+92256 
+10

This question made me think 

 

$$\\cos\;\frac{\pi}{2}=0,\qquad cos\;\frac{-\pi}{2}=0,\qquad and \qquad f(0)=0\\\\$$

 

 

Now since g(x)=f(x)*cosx it follows that g(x)=0 when x  = -pi/2, 0 or pi/2

Hence g(x) must be at least one turning point between  -pi/2 and 0  and at least one between 0 and pi/2

If you are not sure about this, plot the 3 points and think about what the graph will have to look like.

NOW  when g(x) turns the gradient of the tangent must be 0 SO g'(x)=0 

HENCE

1 and 3 must both be correct. 

Melody  Feb 22, 2015
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3+0 Answers

 #2
avatar+92256 
+10
Best Answer

This question made me think 

 

$$\\cos\;\frac{\pi}{2}=0,\qquad cos\;\frac{-\pi}{2}=0,\qquad and \qquad f(0)=0\\\\$$

 

 

Now since g(x)=f(x)*cosx it follows that g(x)=0 when x  = -pi/2, 0 or pi/2

Hence g(x) must be at least one turning point between  -pi/2 and 0  and at least one between 0 and pi/2

If you are not sure about this, plot the 3 points and think about what the graph will have to look like.

NOW  when g(x) turns the gradient of the tangent must be 0 SO g'(x)=0 

HENCE

1 and 3 must both be correct. 

Melody  Feb 22, 2015
 #3
avatar+85958 
+5

Jesus, that was so simple , Melody.......I was trying to do all sorts of "exotic" things to figure this out......LOL!!!.......there's genius in simplicity......!!!

 

CPhill  Feb 22, 2015
 #4
avatar+92256 
+5

Thanks Chris, that is high praise. 

Melody  Feb 23, 2015

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