Given f(x) = <span class="math-display">\[ \begin{cases} |x+1| & \text{if } x < -2 \\ 2x + 3 & \text{if } -2 \leq x < 0 \\ x^2 + 3 & \text{if } 0 \leq x < 3 \\ x^3 - 15 & \text{if } x \geq 3 \end{cases} \]</span>. Then number of point(s) of discontinuity of f(x) is-
Step-by-Step Solution
Key Concept: A piecewise function is continuous at a point if the left-hand limit, right-hand limit, and function value all exist and are equal at that point. We must check continuity at the boundary points where the function definition changes: x = -2, x = 0, and x = 3.
Step 1: Identify boundary points
The function $f(x)$ is defined piecewise, with definitions changing at $x = -2$, $x = 0$, and $x = 3$. We must check for continuity at these points. Within each interval, the component functions (absolute value, linear, quadratic, and cubic) are continuous.
Step 2: Check continuity at $x = -2$
To check continuity at $x = -2$, we evaluate the left-hand limit, the right-hand limit, and the function value at $x = -2$.
The left-hand limit is:
$$ \lim_{x\to-2^-} f(x) = \lim_{x\to-2^-} |x+1| = |-2+1| = |-1| = 1 $$
The right-hand limit is:
$$ \lim_{x\to-2^+} f(x) = \lim_{x\to-2^+} (2x+3) = 2(-2)+3 = -4+3 = -1 $$
The function value at $x = -2$ is:
$$ f(-2) = 2(-2)+3 = -4+3 = -1 $$
Since $\lim_{x\to-2^-} f(x) = 1$ and $\lim_{x\to-2^+} f(x) = -1$, the left-hand limit does not equal the right-hand limit ($1 \neq -1$).
Therefore, the function $f(x)$ is **discontinuous at $x = -2$**.
Step 3: Check continuity at $x = 0$
To check continuity at $x = 0$, we evaluate the left-hand limit, the right-hand limit, and the function value at $x = 0$.
The left-hand limit is:
$$ \lim_{x\to0^-} f(x) = \lim_{x\to0^-} (2x+3) = 2(0)+3 = 3 $$
The right-hand limit is:
$$ \lim_{x\to0^+} f(x) = \lim_{x\to0^+} (x^2+3) = 0^2+3 = 3 $$
The function value at $x = 0$ is:
$$ f(0) = 0^2+3 = 3 $$
Since $\lim_{x\to0^-} f(x) = 3$, $\lim_{x\to0^+} f(x) = 3$, and $f(0) = 3$, all three values are equal.
Therefore, the function $f(x)$ is **continuous at $x = 0$**.
Step 4: Check continuity at $x = 3$
To check continuity at $x = 3$, we evaluate the left-hand limit, the right-hand limit, and the function value at $x = 3$.
The left-hand limit is:
$$ \lim_{x\to3^-} f(x) = \lim_{x\to3^-} (x^2+3) = 3^2+3 = 9+3 = 12 $$
The right-hand limit is:
$$ \lim_{x\to3^+} f(x) = \lim_{x\to3^+} (x^3-15) = 3^3-15 = 27-15 = 12 $$
The function value at $x = 3$ is:
$$ f(3) = 3^3-15 = 27-15 = 12 $$
Since $\lim_{x\to3^-} f(x) = 12$, $\lim_{x\to3^+} f(x) = 12$, and $f(3) = 12$, all three values are equal.
Therefore, the function $f(x)$ is **continuous at $x = 3$**.
Step 5: Count discontinuities
Based on the analysis, the function $f(x)$ is discontinuous only at $x = -2$.
The number of points of discontinuity of $f(x)$ is 1.
Correct Answer: D