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.