Question

The number of points where the function

$f(x) = \begin{cases} |2x^2 - 3x - 7| & \text{if } x \leq -1 \\ [4x^2 - 1] & \text{if } -1 < x < 1 \\ |x+1| + |x-2| & \text{if } x \geq 1 \end{cases}$

$[t]$ denotes the greatest integer $\leq t$, is discontinuous is.

• A. 0
• B. 1
• C. 7
• D. 8

Answer 1

Answer: (C)

7

Answer 2

The function $f(x)$ is defined in three pieces. We analyze the continuity within each piece and at the points where the definition changes ($x=-1$ and $x=1$).

\textbf{1. For $x \leq -1$, $f(x) = |2x^2 - 3x - 7|$:} This is a composition of continuous functions (a polynomial and the absolute value function), so it is continuous everywhere within its domain. No discontinuities arise from this piece.

\textbf{2. For $-1 < x < 1$, $f(x) = [4x^2 - 1]$:} The greatest integer function $[t]$ is discontinuous when its argument $t$ is an integer. Let $g(x) = 4x^2 - 1$. We need to find values of $x$ in the interval $(-1, 1)$ for which $g(x)$ is an integer. Let $4x^2 - 1 = k$, where $k$ is an integer. $4x^2 = k+1 \implies x^2 = \frac{k+1}{4}$. For $x \in (-1, 1)$, we have $0 \leq x^2 < 1$. Therefore, $0 \leq \frac{k+1}{4} < 1$, which implies $0 \leq k+1 < 4$, so $-1 \leq k < 3$. The possible integer values for $k$ are $-1, 0, 1, 2$.

\begin{itemize} \item If $k = -1$: $x^2 = 0 \implies x = 0$. At $x=0$, $g(0) = -1$. $\lim_{x \to 0^-} [4x^2 - 1] = [-1^+] = -1$. $\lim_{x \to 0^+} [4x^2 - 1] = [-1^+] = -1$. $f(0) = [-1] = -1$. Since the left-hand limit, right-hand limit, and the function value are equal, $f(x)$ is \textbf{continuous} at $x=0$.

\item If $k = 0$: $x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2}$.
At $x = 1/2$: $g(1/2) = 0$.
$\lim_{x \to (1/2)^-} [4x^2 - 1] = [0^-] = -1$.
$\lim_{x \to (1/2)^+} [4x^2 - 1] = [0^+] = 0$.
$f(1/2) = [0] = 0$.
Since $\lim_{x \to (1/2)^-} f(x) \neq \lim_{x \to (1/2)^+} f(x)$, $f(x)$ is \textbf{discontinuous} at $x = 1/2$.
At $x = -1/2$: $g(-1/2) = 0$.
$\lim_{x \to (-1/2)^-} [4x^2 - 1] = [0^+] = 0$.
$\lim_{x \to (-1/2)^+} [4x^2 - 1] = [0^-] = -1$.
$f(-1/2) = [0] = 0$.
Since $\lim_{x \to (-1/2)^-} f(x) \neq \lim_{x \to (-1/2)^+} f(x)$, $f(x)$ is \textbf{discontinuous} at $x = -1/2$.

\item If $k = 1$: $x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$.
At $x = 1/\sqrt{2}$: $g(1/\sqrt{2}) = 1$.
$\lim_{x \to (1/\sqrt{2})^-} [4x^2 - 1] = [1^-] = 0$.
$\lim_{x \to (1/\sqrt{2})^+} [4x^2 - 1] = [1^+] = 1$.
$f(1/\sqrt{2}) = [1] = 1$.
$f(x)$ is \textbf{discontinuous} at $x = 1/\sqrt{2}$.
At $x = -1/\sqrt{2}$: $g(-1/\sqrt{2}) = 1$.
$\lim_{x \to (-1/\sqrt{2})^-} [4x^2 - 1] = [1^+] = 1$.
$\lim_{x \to (-1/\sqrt{2})^+} [4x^2 - 1] = [1^-] = 0$.
$f(-1/\sqrt{2}) = [1] = 1$.
$f(x)$ is \textbf{discontinuous} at $x = -1/\sqrt{2}$.

\item If $k = 2$: $x^2 = \frac{3}{4} \implies x = \pm \frac{\sqrt{3}}{2}$.
At $x = \sqrt{3}/2$: $g(\sqrt{3}/2) = 2$.
$\lim_{x \to (\sqrt{3}/2)^-} [4x^2 - 1] = [2^-] = 1$.
$\lim_{x \to (\sqrt{3}/2)^+} [4x^2 - 1] = [2^+] = 2$.
$f(\sqrt{3}/2) = [2] = 2$.
$f(x)$ is \textbf{discontinuous} at $x = \sqrt{3}/2$.
At $x = -\sqrt{3}/2$: $g(-\sqrt{3}/2) = 2$.
$\lim_{x \to (-\sqrt{3}/2)^-} [4x^2 - 1] = [2^+] = 2$.
$\lim_{x \to (-\sqrt{3}/2)^+} [4x^2 - 1] = [2^-] = 1$.
$f(-\sqrt{3}/2) = [2] = 2$.
$f(x)$ is \textbf{discontinuous} at $x = -\sqrt{3}/2$.

\end{itemize} The points of discontinuity in the interval $(-1, 1)$ are $x = \pm \frac{1}{2}, \pm \frac{1}{\sqrt{2}}, \pm \frac{\sqrt{3}}{2}$. There are \textbf{6} such points.

\textbf{3. For $x \geq 1$, $f(x) = |x+1| + |x-2|$:} This function is a sum of absolute value functions, which is continuous for all $x$. No discontinuities arise from this piece.

\textbf{4. Checking boundary points:}

\textbf{At $x = -1$:} Left-hand limit: $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} |2x^2 - 3x - 7| = |2(-1)^2 - 3(-1) - 7| = |2+3-7| = |-2| = 2$. Right-hand limit: $\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} [4x^2 - 1]$. As $x \to -1^+$, $x^2 \to 1^-$, so $4x^2 - 1 \to 3^-$. Thus, $\lim_{x \to -1^+} [4x^2 - 1] = [3^-] = 2$. Function value: $f(-1) = |2(-1)^2 - 3(-1) - 7| = 2$. Since $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) = f(-1) = 2$, $f(x)$ is \textbf{continuous} at $x = -1$.

\textbf{At $x = 1$:} Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} [4x^2 - 1]$. As $x \to 1^-$, $x^2 \to 1^-$, so $4x^2 - 1 \to 3^-$. Thus, $\lim_{x \to 1^-} [4x^2 - 1] = [3^-] = 2$. Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (|x+1| + |x-2|)$. For $x > 1$ and close to 1, $|x+1| = x+1$ and $|x-2| = -(x-2) = 2-x$. $\lim_{x \to 1^+} (x+1 + 2-x) = \lim_{x \to 1^+} 3 = 3$. Function value: $f(1) = |1+1| + |1-2| = |2| + |-1| = 2 + 1 = 3$. Since $\lim_{x \to 1^-} f(x) = 2$ and $\lim_{x \to 1^+} f(x) = 3$, $f(x)$ is \textbf{discontinuous} at $x = 1$.

\textbf{Conclusion:} The points of discontinuity are the 6 points in $(-1, 1)$ where $[4x^2 - 1]$ is discontinuous ($\pm 1/2, \pm 1/\sqrt{2}, \pm \sqrt{3}/2$), plus the point $x=1$. The total number of points of discontinuity is $6 + 1 = 7$.