Question

Determine the continuity of the function $f(x)$ defined as follows:

$f(x) = \begin{cases} x^2+1 & \text{if } x < 1 \\ 2x+3 & \text{if } x \geq 1 \end{cases}$

• A. Discontinuous at x=0
• B. Continuous everywhere
• C. Discontinuous at x=1
• D. Answer Choice 4
• E. Continuous at x=1 but discontinuous elsewhere

Answer 1

Answer: (C)

Discontinuous at x=1

Answer 2

To determine the continuity of the function $f(x)$, we need to examine its behavior at different points. The function is defined piecewise:

$f(x) = \begin{cases} x^2+1 & \text{if } x < 1 \\ 2x+3 & \text{if } x \geq 1 \end{cases}$

1. Continuity in the intervals:

• For $x < 1$, $f(x) = x^2+1$. This is a polynomial function, which is continuous for all real numbers. Therefore, $f(x)$ is continuous for all $x < 1$.
• For $x > 1$, $f(x) = 2x+3$. This is also a polynomial (linear) function, which is continuous for all real numbers. Therefore, $f(x)$ is continuous for all $x > 1$.

2. Continuity at the point where the definition changes ($x=1$):

For a function to be continuous at a point $x=a$, the following condition must be met: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$

Let's check this for $x=1$:

• Value of the function at $x=1$: Since $x \geq 1$, we use the second part of the definition: $f(1) = 2(1) + 3 = 2 + 3 = 5$.

• Left-hand limit at $x=1$ (LHL): For $x < 1$, $f(x) = x^2+1$. $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2+1) = (1)^2+1 = 1+1 = 2$.

• Right-hand limit at $x=1$ (RHL): For $x \geq 1$, $f(x) = 2x+3$. $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x+3) = 2(1)+3 = 2+3 = 5$.

Conclusion:

We observe that the left-hand limit (2) is not equal to the right-hand limit (5). Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$, the limit $\lim_{x \to 1} f(x)$ does not exist. Therefore, the function $f(x)$ is discontinuous at $x=1$.

The function is continuous for all $x \neq 1$ and discontinuous at $x=1$.