Question

attempt 1: 10 attempts remaining.
determine the x-values where f(x) is discontinuous.

f(x) =
\

$$\begin{cases} x + 1 & \\text{if } x \\leq 1 \\\\ 1/x & \\text{if } 1 < x < 3 \\\\ \\sqrt{x - 3} & \\text{if } x \\geq 3 \\end{cases}$$

if there are none, enter none. if there are multiple discontinuities, separate the values with commas.

x =

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Explanation:

Step1: Check continuity at \( x = 1 \)

For \( x \leq 1 \), \( f(x)=x + 1 \). So \( f(1)=1 + 1 = 2 \).
For \( 1 < x < 3 \), \( f(x)=\frac{1}{x} \). The left - hand limit as \( x\to1^{+} \) is \( \lim_{x\to1^{+}}\frac{1}{x}=1 \).
Since \( f(1) = 2
eq\lim_{x\to1^{+}}f(x)=1 \), the function is discontinuous at \( x = 1 \).

Step2: Check continuity at \( x = 3 \)

For \( 1 < x < 3 \), \( f(x)=\frac{1}{x} \). The left - hand limit as \( x\to3^{-} \) is \( \lim_{x\to3^{-}}\frac{1}{x}=\frac{1}{3} \).
For \( x\geq3 \), \( f(x)=\sqrt{x - 3} \). So \( f(3)=\sqrt{3 - 3}=0 \). And the right - hand limit as \( x\to3^{+} \) is \( \lim_{x\to3^{+}}\sqrt{x - 3}=0 \). But \( \lim_{x\to3^{-}}f(x)=\frac{1}{3}
eq f(3) = 0 \), so the function is discontinuous at \( x = 3 \).

Step3: Check other points

• For \( x<1 \), \( f(x)=x + 1 \) is a polynomial, so it is continuous everywhere on \( (-\infty,1) \).
• For \( 1 < x < 3 \), \( f(x)=\frac{1}{x} \) is a rational function (and \( x

eq0 \) in this interval), so it is continuous on \( (1,3) \).

• For \( x>3 \), \( f(x)=\sqrt{x - 3} \) is a square - root function and is continuous on \( (3,\infty) \) (since the expression inside the square root \( x - 3>0 \) for \( x>3 \)).

Answer:

1, 3