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

1 Answer
Andrea S.
Dec 1, 2016

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]}

Related questions
See all questions in Identifying Stationary Points (Critical Points) for a Function
Impact of this question
2085 views around the world
You can reuse this answer
Creative Commons License