Question

If $g\left( x \right) = x$ if $x < 0$, ${x^2}$ if $0 \leqslant x \leqslant 1$, and ${x^3}$ if $x > 1$ . How do you show that g is continuous on all real numbers ? $$

Answer 1

Hint : In the given question, we are required to find whether a function given to us is continuous or not. The problem involves concepts of continuity and differentiability and can be solved easily with the knowledge of conditions for a function to be continuous. In the question, we are given a piece by piece defined function. This means the behavior of the graph of the function changes for different intervals of x.

Complete step by step solution:
Let us consider a real number c, where $c \in ( - \infty ,0)$ .
For $x < 0$, $g\left( x \right) = x$ .
So, $\mathop {\lim }\limits_{x \to c} g(x) = \mathop {\lim }\limits_{x \to c} x$
$\Rightarrow \mathop {\lim }\limits_{x \to c} g(x) = c$
Also, $g(c) = c$
Thus, $\mathop {\lim }\limits_{x \to c} g(x) = g(c) = c$
Hence the given function g(x) is continuous on $( - \infty ,0)$ .

Now, evaluating limit at $x = 0$
$\mathop {\lim }\limits_{x \to 0} g(x) = \mathop {\lim }\limits_{x \to 0} x = 0$
Also, $g(0) = 0$
Thus, $\mathop {\lim }\limits_{x \to 0} g(x) = g(0) = 0$
Hence the given function g(x) is continuous at x equals zero.

Now let us consider a real number d, where $d \in (0,1)$
$\mathop {\lim }\limits_{x \to d} g(d) = \mathop {\lim }\limits_{x \to d} {x^2}$
Or, $\mathop {\lim }\limits_{x \to d} g(d) = {d^2}$
Also, $g(d) = {d^2}$
Thus, $\mathop {\lim }\limits_{x \to d} g(d) = g(d) = {d^2}$
Hence the given function g(x) is continuous on $(0,1)$.
For $0 \leqslant x \leqslant 1$, $g\left( x \right) = {x^2}$ .
So, evaluating limit at $x = 1$,
$\mathop {\lim }\limits_{x \to 1} g(x) = \mathop {\lim }\limits_{x \to 1} {x^2}$
Also, $g(1) = 1$
Thus, $\mathop {\lim }\limits_{x \to 1} g(x) = g(1) = 1$
Hence the given function g(x) is continuous at x equals one.
Now, consider a real number e, $e \in (1,\infty )$

For $x > 1$, $g\left( x \right) = {x^3}$ ,
$\mathop {\lim }\limits_{x \to e} g(x) = \mathop {\lim }\limits_{x \to e} {x^3}$
Or, $\mathop {\lim }\limits_{x \to e} g(x) = {e^3}$
Also, $g(e) = {e^3}$
Thus, $\mathop {\lim }\limits_{x \to e} g(x) = g(e) = {e^3}$
Thus the given function g(x) is continuous on $(1,\infty )$
Therefore, we conclude that the given function g(x) is continuous on all real numbers.

Note : A function is said to be continuous if the graph of the function does not have any empty spaces in between or can be drawn on a graph paper without lifting up the pencil or pen. Functions that are defined piecewise behave differently for different sets of values of x.