Let $f: [-2, 2] o \mathbb{R}$ be defined as $f(x) = \begin{cases} (x+1)2^{-(\frac{1}{|x|} + \frac{

Question

Let $f: [-2, 2] 	o \mathbb{R}$ be defined as $f(x) = \begin{cases} (x+1)2^{-(\frac{1}{|x|} + \frac{1}{x})} & 	ext{if } x 
eq 0 \ 0 & 	ext{if } x = 0 \end{cases}$. Which of the following statements is/are CORRECT?

Professor PC Thomas Classes · Answer

First, simplify the function. For $x < 0$, $|x| = -x$, so the exponent is $-(-1/x + 1/x) = 0$. Thus, $f(x) = (x+1)2^0 = x+1$. For $x > 0$, $|x| = x$, so the exponent is $-(1/x + 1/x) = -2/x$. Thus, $f(x) = (x+1)2^{-2/x}$. Now, evaluate the limits as $x$ approaches 0. $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (x+1) = 1$. $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} (x+1)2^{-2/x} = (1) \cdot 2^{-\infty} = 0$. Since the left-hand and right-hand limits are finite but unequal, $f(x)$ has a jump discontinuity at $x=0$. This makes (A) and (B) correct. Next, let us evaluate the endpoints: $f(-2) = -2+1 = -1$ and $f(2) = (2+1)2^{-2/2} = 3/2$. The range of the left branch $[-2, 0)$ is $[-1, 1)$. For the right branch $(0, 2]$, the function $f(x) = (x+1)2^{-2/x}$ is strictly increasing (as its derivative is positive), giving a range of $(0, 3/2]$. Combining these with $f(0) = 0$, the overall range on $[-2, 2]$ is $[-1, 1) \cup \{0\} \cup (0, 3/2] = [-1, 3/2]$, which is exactly $[f(-2), f(2)]$. Thus, the function attains all intermediate values despite being discontinuous, making (D) correct and (C) incorrect.