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