Let f(x)={[ln(-x)+3," for "x < -3],[ln(-x+3)," for "-3 <= x < 3]:} Is f continuous at x=-3 ? Choose 1 answer: (A) Yes (B) No

Check Continuity at -3: To determine if the function f(x) is continuous at x=-3, we need to check if the left-hand limit, the right-hand limit, and the value of the function at x=-3 are all equal.Calculate Left-hand Limit: First, we calculate the left-hand limit as x approaches -3 from the left. This means we use the first part of the function definition, f(x) = ln(-x) + 3. lim(x->-3^-) f(x) = lim(x->-3^-) (ln(-x) + 3)Calculate Right-hand Limit: Substitute x with a value slightly less than -3, say -3.01, into the function to get an idea of the left-hand limit. f(-3.01) = ln(-(-3.01)) + 3 = ln(3.01) + 3Left-hand Limit Calculation: Since the natural logarithm function is continuous and ln(3) is a finite number, as x approaches -3 from the left, ln(-x) approaches ln(3), and thus the left-hand limit is ln(3) + 3. lim(x->-3^-) f(x) = ln(3) + 3Right-hand Limit Calculation: Next, we calculate the right-hand limit as x approaches -3 from the right. This means we use the second part of the function definition, f(x) = ln(-x+3). lim(x->-3^+) f(x) = lim(x->-3^+) (ln(-x+3))Conclusion: Substitute x with a value slightly greater than -3, say -2.99, into the function to get an idea of the right-hand limit. f(-2.99) = ln(-(-2.99) + 3) = ln(0.01)As x approaches -3 from the right, -x+3 approaches 0, and the natural logarithm of a number approaching 0 from the positive side goes to negative infinity. Therefore, the right-hand limit does not exist because ln(0) is undefined. lim(x->-3^+) f(x) = ln(0) = -∞Since the right-hand limit as x approaches -3 is negative infinity, which is not equal to the left-hand limit, the function f(x) is not continuous at x=-3.

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Let

f(x)={[ln(-x)+3," for "x < -3],[ln(-x+3)," for "-3 <= x < 3]:}
Is
f continuous at
x=-3 ?
Choose 1 answer:
(A) Yes
(B)
No

Let $\newline$ $f(x)=\left\{\begin{array}{ll} \ln (-x)+3 & \text { for } x<-3 \\ \ln (-x+3) & \text { for }-3 \leq x<3 \end{array}\right.$ $\newline$ Is $f$ continuous at $x=-3$ ? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) $\mathrm{No}$

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Math Problems

Calculus

Intermediate Value Theorem

Full solution

Q. Let

\newline

f(x)=\left\{\begin{array}{ll} \ln (-x)+3 & \text { for } x<-3 \\ \ln (-x+3) & \text { for }-3 \leq x<3 \end{array}\right.

\newline

f

continuous at

x=-3

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B)

\mathrm{No}

Check Continuity at $-3$ : To determine if the function $f(x)$ is continuous at $x=-3$ , we need to check if the left-hand limit, the right-hand limit, and the value of the function at $x=-3$ are all equal.
Calculate Left-hand Limit: First, we calculate the left-hand limit as $x$ approaches $-3$ from the left. This means we use the first part of the function definition, $f(x) = \ln(-x) + 3$ . $\newline$ $\lim_{x\to-3^-} f(x) = \lim_{x\to-3^-} (\ln(-x) + 3)$
Calculate Right-hand Limit: Substitute $x$ with a value slightly less than $-3$ , say $-3.01$ , into the function to get an idea of the left-hand limit. $\newline$ $f(-3.01) = \ln(-(-3.01)) + 3 = \ln(3.01) + 3$
Left-hand Limit Calculation: Since the natural logarithm function is continuous and $\ln(3)$ is a finite number, as $x$ approaches $-3$ from the left, $\ln(-x)$ approaches $\ln(3)$ , and thus the left-hand limit is $\ln(3) + 3$ . $\newline$ $\lim_{x\to-3^-} f(x) = \ln(3) + 3$
Right-hand Limit Calculation: Next, we calculate the right-hand limit as $x$ approaches $-3$ from the right. This means we use the second part of the function definition, $f(x) = \ln(-x+3)$ . $\newline$ $\lim_{x\to-3^+} f(x) = \lim_{x\to-3^+} (\ln(-x+3))$
Conclusion: Substitute $x$ with a value slightly greater than $-3$ , say $-2.99$ , into the function to get an idea of the right-hand limit. $\newline$ $f(-2.99) = \ln(-(-2.99) + 3) = \ln(0.01)$
Conclusion: Substitute $x$ with a value slightly greater than $-3$ , say $-2.99$ , into the function to get an idea of the right-hand limit. $f(-2.99) = \ln(-(-2.99) + 3) = \ln(0.01)$ As $x$ approaches $-3$ from the right, $-x+3$ approaches $0$ , and the natural logarithm of a number approaching $0$ from the positive side goes to negative infinity. Therefore, the right-hand limit does not exist because $\ln(0)$ is undefined. $\lim_{x\to-3^+} f(x) = \ln(0) = -\infty$
Conclusion: Substitute $x$ with a value slightly greater than $-3$ , say $-2.99$ , into the function to get an idea of the right-hand limit. $f(-2.99) = \ln(-(-2.99) + 3) = \ln(0.01)$ As $x$ approaches $-3$ from the right, $-x+3$ approaches $0$ , and the natural logarithm of a number approaching $0$ from the positive side goes to negative infinity. Therefore, the right-hand limit does not exist because $\ln(0)$ is undefined. $\lim_{x\to-3^+} f(x) = \ln(0) = -\infty$ Since the right-hand limit as $x$ approaches $-3$ is negative infinity, which is not equal to the left-hand limit, the function $-3$ $1$ is not continuous at $-3$ $2$ .