Question

Let $g\left(x\right)=\frac{1}{x^2},x>0;g\left(0\right)=0$

• A. $g$ is a continuous function
• B. $g$ is a bounded function
• C. $g$ is bounded on $[1,\infty ]$
• D. $g$ has a minimum value on$[I,\infty ]$

Answer 1

Answer: (C)

$g$ is bounded on $[1,\infty ]$

The graph for $g\left(x\right)$ is shown in the figure. The value of $g\left(0\right)$ is $0$.
The function $g\left(x\right)$ is continuous everywhere except at $g\left(x\right)=0$.
The function $g\left(x\right)$ is unbounded because it assumes values in the interval $(-\infty ,\infty )$. For a function to be bounded, its values must always lie in a finite interval for all $x$.
The function $g\left(x\right)$ is bounded in the interval $[1,\infty )$ because it assumes values in a finite interval: $(0,1]$ ($\underset{x\to \infty }{lim}g\left(x\right)=0$ and $g\left(1\right)=1$.