Use the definition of derivative

Question

Let g(x) = |cos x|.

(a) Graph g(x) and g'(x) on one graph for −2π ≤ x ≤ 2π

 

(b) find g'(π/2)

 

how would you show that it doesn't exist.

 

(c) g'(0)

 

 

 

Answer 1

Jason,

 

(a) When graphed, the absolute value of cos looks like a bouncing ball because wherever the normal cos x becomes negative you have to make it positive by changing its sign.  

 

Let's make a table of 5 domains of x covering [-2Π, 2Π]

 

x - domain        g(x) = |COS x|          g'(x) 

========  ===========  ==========

[-2Π, -3Π/2]       +  COS x             -  SIN x   

[-3Π/2, -Π/2]      -   COS x            +  SIN x

[-Π/2, +Π/2]      +   COS x            -   SIN x

[+Π/2,+3Π/2]     -   COS x            +  SIN x  

[+3Π/2,+2Π]      +  COS x             -  SIN x

 

 

Note that at x = -2Π, -Π, 0, Π, and 2Π,  g(x) = |COS x| = 1, its maximum value

        and at x = -3Π/2, -Π/2, Π/2, and 3Π/2 , g(x) = 0, its minimum value.  Hence, the bouncing ball.

You can make appropriate comments for g'(x) at these points from the above table.

 

(b) g'(x=Π/2) does not exist because g'(x) is not continuous at x = Π/2 so there is no unique tangent there.

Approaching x = Π/2 from the left (in above table), g'(x) = - SIN x so g'(Π/2) = - 1 as the limit.

But approaching x = Π/2 from the right (in above table), g'(x) = + SIN x so g'(Π/2) = + 1 as the limit.

 

(c)  g'(0) = - SIN (0) = 0