Let $f: [-1, 1] o \mathbb{R}$ be defined as $f(x) = \begin{cases} 2x & ext{if } -1 \le x \le 0 \

Question

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

Professor PC Thomas Classes · Answer

To check for continuity at $x = 0$, we evaluate the left-hand limit (LHL) and right-hand limit (RHL). LHL = $\lim_{x 	o 0^-} 2x = 0$. RHL = $\lim_{x 	o 0^+} (x + 1/2) = 1/2$. Since LHL $
eq$ RHL and both are finite, $f(x)$ has a jump discontinuity (first kind) at $x = 0$. This makes (A) correct. To find the extrema, we analyze the intervals separately. On $[-1, 0]$, $f(x) = 2x$ is strictly increasing, with a minimum of $f(-1) = -2$ and a maximum of $f(0) = 0$. On $(0, 1]$, $f(x) = x + 1/2$ is strictly increasing, bounded below by $1/2$ (not attained) and reaching a maximum of $f(1) = 3/2$. Comparing the two pieces, the absolute minimum is $-2$ (attained at $x = -1$) and the absolute maximum is $3/2$ (attained at $x = 1$). This makes (B) correct. The overall range is the union of the ranges of the individual branches, which is $[-2, 0] \cup (1/2, 3/2]$, making (C) correct. (D) is incorrect because the Extreme Value Theorem explicitly requires continuity on a closed interval; this function merely demonstrates that continuity is a sufficient, but not necessary, condition for the existence of global extrema.