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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability2 Marks
Question
Discuss the continuity of the function f given by
$f(x)=\left\{\begin{array}{ll} {x,} & {\text { if } x \geq 0} \\ {x^{2},} & {\text { if } x<0} \end{array}\right.$
✓

Answer

Clearly, the function is defined at every real number. The graph of the function is given in figure. By inspection, it seems prudent to partition the domain of definition of f into three disjoint subsets of the real line.

Let $D_1 =\{x  \in  R : x < 0\}, D_2 = \{0\}$ and $​D​​​​​​_3 = \{x \in R : x > 0\}$​​​​​​​
Case 1: At any point in $D_1$ , we have $f(x) = x^2$ and it is easy to see that it is continuous there
Case 2: At any point in $D_3$ , we have $f(x) = x$ and it is easy to see that it is continuous there
Case 3: Now we analyse the function at $x = 0.$ The value of the function at $0$ is $f(0) = 0.$ The left hand limit of $f$ at $0$ is
$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} {x^2} = {0^2} = 0$
The right-hand limit of $f$ at $0$ is
$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} x = 0$
Thus $\mathop {\lim }\limits_{x \to {0^ + }} f(x)$ = 0 = f(0) and hence f is continuous at 0. This means that f is continuous at every point in its domain and hence, f is a continuous function.

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