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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

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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

#2
+91412
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$$\\cos\;\frac{\pi}{2}=0,\qquad cos\;\frac{-\pi}{2}=0,\qquad and \qquad f(0)=0\\\\$$

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

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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#2
+91412
+10

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

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

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
+80804
+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
+91412
+5

Thanks Chris, that is high praise.

Melody  Feb 23, 2015

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