How do you find the critical points for #f(x)= (sinx) / (cosx-2)# and the local max and min?

1 Answer
Dec 1, 2016

Answer:

#f(x)# has a maximum in #x=-pi/3+2kpi# and a minimum in #x=pi/3+2xpi#

Explanation:

Critical points are where the first derivative is zero.

#f(x) = frac sinx (cosx-2)#

#f'(x) = frac ( (cosx -2) cosx + sin^2x) ((cosx-2)^2) = frac ( cos^2x -2 cosx + sin^2x) ((cosx-2)^2) = (1-2cosx) / (cosx-2)^2#

So critical points are where:

#1-2cosx=0#

#cosx=1/2#

#x=pi/3+2kpi# and #x=-pi/3+2kpi#

Without calculating the second derivative, we can see that the denominator of #f'(x)# is always positive, so the sign of the derivative is the sign of #1-2cosx#.

So, for #x_k=pi/3+2kpi#, #f'(x_k-epsilon) <0# and #f'(x_k+epsilon) >0# so these points are minimums, while for #x_k=--pi/3+2pi# the opposite is true and these points are maximums.

graph{sinx/(cosx-2) [-10, 10, -5, 5]}