Question

Show that f (x) = | cos x | is a continuous function.

 

Answer 1

The given function is `f(x)=|cos x|`

This function f is defined for every real number and f can be written as the composition of two functions as,

f = g o h, where `g(x)=|x|  and  h(x)=cos  x`

`[∵(goh)(x)=g(h(x))=g(cos  x)=|cos  x|=f(x)]`

It has to be first proved that `g(x)=|x|  and  h(x)=cos x` are continuous functions.

`g(x)=|x| "  can be written as " `

`g(x)=[[-x,if x≤ 0],[x,if x ≥ 0]]`

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

`if c < 0 " then " g (c)= -c and lim\_(x->c)(-x)=-c`

`∴ lim_(x->c)g(x)=g(c)`

So, g is continuous at all points x < 0.

Case II:

`" if c < 0   then "  g (c)= -c and lim\_(x->c)(-x)=-c`

`∴ lim_(x->c)g(x)=g(c)`

So, g is continuous at all points x > 0.

Case III: 

`if  c = 0 , "  then " g(c)=g(0)=0`

`lim_(x->0^-)g(x)=lim_(x->0^-)(-x)=0`

`lim_(x->0^+)g(x)=lim_(x->0^+)(x)=g(0)`

`∴lim_(x->0^+)g(x)=lim_(x->0^+)(x)=g(0)`

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number.
Put x = c + h

If x → c, then h → 0

h (c) = cos c

`lim_(x->0)h(x)=lim_(x->0) cos x`

                   `=lim_(k->0) cos (c+h)`

                   `=lim_(k->0)[cos  c  cos  h-sin  c sin h]`

                   `=lim_(k->0)cos  c cos 0 - sin  c sin 0`

                  `= cos  c xx1 - sin  cxx0`

                  `= cos  c`

`lim_(x->c)h(x)=h(c)`

So, h (x) = cos x is a continuous function.

It is known that for real valued functions g and h,such that (g o h) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then (f o g) is continuous at x = c.

Therefore, `f(x)=(goh)(x)=g(h(x))=g(cos x)=|cos x|` is a continuous function.