Question

Find if the given below function is continuous or discontinuous at the indicated point
$\{\begin{array}{cc}\frac{x^2}{2}\text{if}0\le x\le 1& \\ 2x^2-3x+\frac{3}{2},\text{if}1<x\le 2& \end{array}$
at $x=1$

Answer 1

Use the concept of continuity of a function at a point
$\{\begin{array}{cc}\frac{x^2}{2},if0\le x\le 1& \\ 2x^2-3x+\frac{3}{2},if1<x\le 2& \end{array}$
Find LHL, RHL and value of function (if required) at given point and compare them for checking continuity:
At $x=1$
$L.H.L.={lim}_{x\to 1-}f\left(x\right)={lim}_{h\to 0+}f(1-h)$
$={lim}_{h\to 0+}\frac{(1-h{)}^2}{2}$
${lim}_{h\to 0+}\frac{1+h^2-2h}{2}=\frac{1}{2}...\left(2\right)$
$(a-b{)}^2=a^2+b^2-2ab$
$R.H.L={lim}_{x\to 1+}f\left(x\right)={lim}_{h\to 0+}f(1+h)$
$=li{m}_{h\to 0+}\left[2\right(1+h{)}^2-3(1+h)+\frac{3}{2}]$
$=2-3+\frac{3}{2}=\frac{1}{2}....\left(3\right)$
Now,$f\left(1\right)=\frac{1^2}{2}=\frac{1}{2}...\left(4\right)$
From $\left(2\right),\left(3\right),\left(4\right)$
$L.H.L=R.H.L=f\left(1\right)$
Hence,$f\left(x\right)$ is continuous at $x=1$