On the interval I = [-2, 2],the function f(x) | vedclass

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(A) First,we simplify the expression for $f(x)$ for $x 
eq 0$:If $x > 0$,then $|x| = x$,so $\frac{1}{|x|} + \frac{1}{x} = \frac{1}{x} + \frac{1}{x} =

Vedclass · Accepted Answer

is continuous for all values of $x \in I$

Vedclass · Answer

(A) First,we simplify the expression for $f(x)$ for $x 
eq 0$:If $x > 0$,then $|x| = x$,so $\frac{1}{|x|} + \frac{1}{x} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x}$. Thus,$f(x) = (x + 1) e^{-2/x}$.If $x Given $f(0) = 0$,the function is:$f(x) = \begin{cases} (x + 1) e^{-2/x} & x > 0 \ 0 & x = 0 \ x + 1 & x Now,check continuity at $x = 0$:Left-hand limit: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (x + 1) = 1$.Right-hand limit: $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} (x + 1) e^{-2/x} = (1) \cdot e^{-\infty} = 0$.Since $\lim_{x 	o 0^-} f(x) 
eq \lim_{x 	o 0^+} f(x)$,the function is discontinuous at $x = 0$.Therefore,option $(A)$ is false as the function is not continuous for all $x \in I$.

On the interval $I = [-2, 2]$,the function $f(x) = \begin{cases} (x + 1) e^{-\left[ \frac{1}{|x|} + \frac{1}{x} \right]} & x \neq 0 \\ 0 & x = 0 \end{cases}$ is given. Which one of the following does not hold good?

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