# The points of discontinuity of the function

Question:

The points of discontinuity of the function

$f(x)=\left\{\begin{array}{cl}2 \sqrt{x}, & 0 \leq x \leq 1 \\ 4-2 x, & 1 (a)$x=1, x=\frac{5}{2}$(b)$x=\frac{5}{2}$(c)$x=1, \frac{5}{2}, 4$(d)$x=0,4$Solution: (b)$x=\frac{5}{2}$If$0 \leq x \leq 1$, then$f(x)=2 \sqrt{x}$. Since$f(x)=2 \sqrt{x}$is a polynomial function, it is continuous. Thus,$f(x)$is continuous for every$0 \leq x \leq 1$. If$1

If $\frac{5}{2} \leq x \leq 4$, then $f(x)=2 x-7$. Since $2 x$ is a polynomial function and 7 is continuous function, their difference will also be continuous. Thus, $f(x)$ is continuous for every $\frac{5}{2} \leq x \leq 4$.

Now,

Consider the point $x=1$. Here,

$\lim _{\mathrm{x} \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}(2(\sqrt{1-h}))=2$

$\lim _{\mathrm{x} \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(4-2(1+h))=2$

Also, $f(1)=2 \sqrt{1}=2$

$\Rightarrow \lim _{\mathrm{x} \rightarrow 1^{-}} f(x)=\lim _{\mathrm{x} \rightarrow 1^{+}} f(x)=f(1)$

Thus, $f(x)$ is continuous at $x=1$.

Now,

Consider the point $x=\frac{5}{2}$. Here,

$\lim _{x \rightarrow \frac{5}{2}^{-}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{5}{2}-h\right)=\lim _{h \rightarrow 0}\left(4-2\left(\frac{5}{2}-h\right)\right)=-1$

$\lim _{x \rightarrow \frac{5}{2}^{+}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{5}{2}+h\right)=\lim _{h \rightarrow 0}\left(2\left(\frac{5}{2}-h\right)-7\right)=-2$

$\Rightarrow \lim _{x \rightarrow \frac{5}{2}^{+}} f(x) \neq \lim _{x \rightarrow \frac{5}{2}^{-}} f(x)$

Thus, $f(x)$ is discontinuous at $x=\frac{5}{2}$.