Is #f(x)=cosx+tanx-sinx# increasing or decreasing at #x=pi/6#?

1 Answer
Mar 19, 2018

Based on the sign of the derivative, we can see that #f(x)# is decreasing at #x=pi/6#

Explanation:

We will check if the slope of the curve at #x=pi/6# is positive, negative, or zero. This will tell us if the function is increasing (pos.), decreasing (neg.), or steady (zero).

We will find the slope of the curve by taking the derivative of #f(x)#. Since all the terms are additive, we can take the derivatives individually:

#(dy)/(dx)f(x)=f'(x)=d/(dx)cos(x)+d/(dx)tanx-d/(dx)sinx#

#d/(dx)cos(x)=-sinx#

#d/(dx)tanx=sec^2x#

#d/(dx)sinx=cosx#

#f'(x)=-sinx+sec^2x-cosx = sec^2x-(sinx+cosx)#

now, we plug in our x-value, #pi/6#, to find the slope:

#f'(pi/6)=sec^2(pi/6)-(sin(pi/6)+cos(pi/6))#

#f'(pi/6)=1.bar(333)-(0.5+0.8660254)#

#f'(pi/6)=1.bar(333)-(1.366025404)#

#color(red)(f'(pi/6)=-0.03269)#

We can see that the slope is negative, meaning that the function is #color(red)("decreasing")#