Question

determine where f is discontinuous.

f(x) = \

$$\begin{cases} \\sqrt{-x} & \\text{if } x < 0 \\\\ 7 - x & \\text{if } 0 \\leq x < 7 \\\\ (7 - x)^2 & \\text{if } x > 7 \\end{cases}$$

\bigcirc -7 only
\bigcirc 0 only
\bigcirc 0 and -7
\bigcirc 7 only
\bigcirc 0 and 7
\bigcirc no correct answer choice is given.

Explanation:

Step1: Check continuity at \( x = 0 \)

- Left - hand limit (LHL) as \( x\to0^{-} \):
- For \( x\lt0 \), \( f(x)=\sqrt{-x} \). So, \( \lim_{x\to0^{-}}f(x)=\lim_{x\to0^{-}}\sqrt{-x} \). Substitute \( x = 0 \) (approaching from the left), we get \( \sqrt{- 0}=0 \).
- Right - hand limit (RHL) as \( x\to0^{+} \):
- For \( 0\leq x\lt7 \), \( f(x)=7 - x \). So, \( \lim_{x\to0^{+}}f(x)=\lim_{x\to0^{+}}(7 - x) \). Substitute \( x = 0 \), we get \( 7-0 = 7 \).
- Since \( \lim_{x\to0^{-}}f(x)=0 eq\lim_{x\to0^{+}}f(x)=7 \), the function is discontinuous at \( x = 0 \).

Step2: Check continuity at \( x = 7 \)

- Left - hand limit (LHL) as \( x\to7^{-} \):
- For \( 0\leq x\lt7 \), \( f(x)=7 - x \). So, \( \lim_{x\to7^{-}}f(x)=\lim_{x\to7^{-}}(7 - x) \). Substitute \( x = 7 \) (approaching from the left), we get \( 7 - 7=0 \).
- Right - hand limit (RHL) as \( x\to7^{+} \):
- For \( x\gt7 \), \( f(x)=(7 - x)^{2} \). So, \( \lim_{x\to7^{+}}f(x)=\lim_{x\to7^{+}}(7 - x)^{2} \). Substitute \( x = 7 \), we get \( (7 - 7)^{2}=0 \).
- Now, check the value of the function at \( x = 7 \). But the function is not defined at \( x = 7 \) (the middle piece is \( 0\leq x\lt7 \) and the right - hand piece is \( x\gt7 \)). However, for continuity, we need \( \lim_{x\to a}f(x)=f(a) \) (if \( f(a) \) is defined) or just the existence of the limit and the function's behavior around \( a \). But here, \( \lim_{x\to7^{-}}f(x)=\lim_{x\to7^{+}}f(x) = 0 \). But let's check the function's definition: the middle interval is \( 0\leq x\lt7 \) and the right is \( x\gt7 \), so at \( x = 7 \), the function has a "jump" in the sense of the domain? Wait, no, the left - hand limit and right - hand limit are equal. Wait, maybe I made a mistake. Wait, the function at \( x = 7 \): the left - hand side is \( 7 - x \) (approaching 7 from the left, \( x\lt7 \)), the right - hand side is \( (7 - x)^{2} \) (approaching 7 from the right, \( x\gt7 \)). The limit as \( x\to7 \) is 0 (both left and right limits are 0). But the function is not defined at \( x = 7 \)? Wait, no, the middle piece is \( 0\leq x\lt7 \), so \( f(7) \) is not defined in the middle piece, and the right - hand piece is \( x\gt7 \), so \( f(7) \) is not defined. But for a function to be continuous at a point, the point must be in the domain. Wait, maybe the function is defined at \( x = 0 \)? Wait, the middle piece is \( 0\leq x\lt7 \), so \( f(0)=7 - 0 = 7 \). Wait, in step 1, we saw that the left - hand limit at 0 is 0 and the right - hand limit (and the function value at 0) is 7. So the function is discontinuous at 0. At \( x = 7 \), the left - hand limit is \( \lim_{x\to7^{-}}(7 - x)=0 \), the right - hand limit is \( \lim_{x\to7^{+}}(7 - x)^{2}=0 \). But the function is not defined at \( x = 7 \), but the limit exists. However, the function has a discontinuity at \( x = 0 \) (since left and right limits are not equal) and let's re - check \( x = 0 \): the left - hand limit is 0, the function value at \( x = 0 \) is \( f(0)=7 - 0 = 7 \) (from the middle piece, since \( 0\leq x\lt7 \) includes \( x = 0 \)). So \( \lim_{x\to0^{-}}f(x)=0 eq f(0)=7 \), so discontinuous at 0. At \( x = 7 \), the left - hand limit is 0, the right - hand limit is 0, but the function is not defined at \( x = 7 \), but actually, the middle interval is \( 0\leq x\lt7 \) and the right is \( x\gt7 \), so at \( x = 7 \), there is a point where the function changes its formula, but the left and right limits are equal. Wait, maybe the problem is that at \( x = 0 \), the left and right limits are not equal, and at \( x = 7 \), let's check the function values. Wait, maybe I messed up. Let's re - do \( x = 0 \):

Left - hand limit (x approaches 0 from the left, \( x\lt0 \)): \( f(x)=\sqrt{-x} \), so as \( x\to0^{-} \), \( -x\to0^{+} \), so \( \sqrt{-x}\to\sqrt{0}=0 \).

Right - hand limit (x approaches 0 from the right, \( x\gt0 \) but in \( 0\leq x\lt7 \)): \( f(x)=7 - x \), so as \( x\to0^{+} \), \( 7 - x\to7 \).

Function value at \( x = 0 \): from the middle piece \( 0\leq x\lt7 \), \( f(0)=7 - 0 = 7 \). So \( \lim_{x\to0^{-}}f(x)=0 eq f(0)=7 \), so discontinuous at 0.

At \( x = 7 \):

Left - hand limit (x approaches 7 from the left, \( x\lt7 \)): \( f(x)=7 - x \), so \( \lim_{x\to7^{-}}f(x)=7 - 7 = 0 \).

Right - hand limit (x approaches 7 from the right, \( x\gt7 \)): \( f(x)=(7 - x)^{2} \), so \( \lim_{x\to7^{+}}f(x)=(7 - 7)^{2}=0 \).

Now, is the function defined at \( x = 7 \)? The middle piece is \( 0\leq x\lt7 \) (does not include 7), the right - hand piece is \( x\gt7 \) (does not include 7). So the function is not defined at \( x = 7 \). But for continuity, a function is continuous at a point \( a \) if:

1. \( f(a) \) is defined.

2. \( \lim_{x\to a}f(x) \) exists.

3. \( \lim_{x\to a}f(x)=f(a) \).

At \( x = 7 \), condition 1 is not met (function is not defined at \( x = 7 \)), but the limit exists (left and right limits are equal). However, at \( x = 0 \), condition 3 is not met (limit from left is 0, \( f(0)=7 \)).

Wait, but let's check the answer options. The options are about 0, 7, - 7. Let's check \( x=-7 \):

For \( x\lt0 \), \( f(x)=\sqrt{-x} \). At \( x=-7 \), \( f(-7)=\sqrt{-(-7)}=\sqrt{7} \). The limit as \( x\to - 7 \): since for \( x\lt0 \), the function is \( \sqrt{-x} \), which is continuous for \( x\lt0 \) (as a square - root function of a non - negative number, and \( -x\) is positive when \( x\lt0 \)). So \( \lim_{x\to - 7}f(x)=\sqrt{-(-7)}=\sqrt{7}=f(-7) \), so the function is continuous at \( x=-7 \).

So the function is discontinuous at \( x = 0 \) and \( x = 7 \).

Answer:

O 0 and 7