Is #f(x)= 2xsinx # increasing or decreasing at #x=pi/3 #?

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Answer 1 · 2015-12-06T16:08:24

Increasing.

If #f'(pi/3)<0#, the function is decreasing when #x=pi/3#.

If #f'(pi/3)>0#, the function is increasing when #x=pi/3#.

Finding #f'(x)# will require the product rule:

#f'(x)=sinxd/dx[2x]+2xd/dx[sinx]#

Find each derivative separately.

#d/dx[2x]=2#

#d/dx[sinx]=cosx#

Plug them back in.

#f'(x)=2sinx+2xcosx#

#f'(pi/3)=2sin(pi/3)+2(pi/3)cos(pi/3)#

While we could continue to simplify this, it is clear that none of these values will be negative.

Thus #f'(pi/3)>0# and #f(x)# is increasing when #x=pi/3#.