Briselda I.

asked • 12/10/19# Let f(x)= tanx(|pi/2 times cosx|) defined on (-pi,pi). Find the points of discontinuity.

## 1 Expert Answer

John C. answered • 12/10/19

Experienced tutor in trigonometry, precal, and calculus

First, let's rewrite f(x) in terms of sine and cosine:

f(x) = tan(|π/2*cos(x)|) = sin(|π/2*cos(x)|)/cos(|π/2*cos(x)|)

f(x) will be undefined where the denominator is equal to zero, so we will set the cosine portion equal to 0:

cos(|π/2*cos(x)|) = 0

Cosine is only equal to 0 on the open interval (-pi,pi) at π/2 and -π/2. However, with the absolute value in place, we only need the positive value of π/2 so we set the argument (i.e. the part inside the cosine function) equal to π/2:

|π/2*cos(x)| = π/2

To get rid of the absolute value bars, we would now set the portion inside the absolute value bars equal to both π/2 and -π/2:

π/2*cos(x) = π/2 and π/2*cos(x) = - π/2

If we isolate the cosine function in both equations by dividing both sides of both equations by π/2, we would get:

cos(x) = 1 and cos(x) = - 1

The only values of x that satisfy these equations are x = 0, π, and - π. However, π and - π are not contained in the open interval (-π, π). So, the only solution to our equation is x =0.

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Paul M.

12/10/19