Show that the function defined by is a continuous function.
Question:

Show that the function defined by $f(x)=|\cos x|$ is a continuous function.

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,

$f=g \circ h$, where $g(x)=|x|$ and $h(x)=\cos x$

$[\because(g o h)(x)=g(h(x))=g(\cos x)=|\cos x|=f(x)]$

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

$g(x)= \begin{cases}-x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{cases}$

Clearlyg is defined for all real numbers.

Let c be a real number.

 

Case I:

If $c<0$, then $g(c)=-c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(-x)=-c$

$\therefore \lim _{x \rightarrow c} g(x)=g(c)$

Thereforeg is continuous at all points x, such that x < 0

 

Case II:

If $c>0$, then $g(c)=c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c} x=c$

$\therefore \lim _{x \rightarrow c} g(x)=g(c)$

Thereforeg is continuous at all points x, such that x > 0

 

Case III:

If $c=0$, then $g(c)=g(0)=0$

$\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0$

$\lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0$

$\therefore \lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=g(0)$

Thereforeg is continuous at x = 0

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

(x) = cos x

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

Let be a real number. Put x = c + h

If $x \rightarrow c$, then $h \rightarrow 0$

(c) = cos c

$\begin{aligned} \lim _{x \rightarrow c} h(x) &=\lim _{x \rightarrow c} \cos x \\ &=\lim _{h \rightarrow 0} \cos (c+h) \\ &=\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h] \\ &=\lim _{h \rightarrow 0} \cos c \cos h-\lim _{h \rightarrow 0} \sin c \sin h \\ &=\cos c \cos 0-\sin c \sin 0 \\ &=\cos c \times 1-\sin c \times 0 \\ &=\cos c \end{aligned}$

$\therefore \lim _{x \rightarrow c} h(x)=h(c)$

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

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

Therefore, $f(x)=(g \circ h)(x)=g(h(x))=g(\cos x)=|\cos x|$ is a continuous function.

 

 

 

 

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