Let g(x)={[(1)/(cos(x))," for "-(pi)/(2) = 0]:} Is g continuous at x=0 ? Choose 1 answer: (A) Yes (B) No

Check Left-hand Limit: To determine if g(x) is continuous at x=0, we need to check if the left-hand limit, the right-hand limit, and the value of the function at x=0 are all equal.Calculate Right-hand Limit: First, let's find the left-hand limit as x approaches 0 from the negative side (x -> 0^-). For x in the interval -(pi)/2 0^-) 1/cos(x) = 1/cos(0) = 1/1 = 1.Evaluate Function at x=0: Next, we need to find the right-hand limit as x approaches 0 from the positive side (x -> 0^+). For x >= 0, g(x) is defined as cos(x+pi). So we need to calculate the limit of cos(x+pi) as x approaches 0 from the right. lim (x -> 0^+) cos(x+pi) = cos(0+pi) = cos(pi) = -1.Determine Continuity at x=0: Now, we need to check the value of the function at x=0. Since x=0 is included in the interval for the second piece of the function, g(x) for x >= 0, we use the second piece to evaluate g(0). g(0) = cos(0+pi) = cos(pi) = -1.We have found that the left-hand limit as x approaches 0 is 1, the right-hand limit as x approaches 0 is -1, and the value of the function at x=0 is -1. Since the left-hand limit does not equal the right-hand limit, the function g(x) is not continuous at x=0.

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Let

g(x)={[(1)/(cos(x))," for "-(pi)/(2) < x < 0],[cos(x+pi)," for "x >= 0]:}
Is
g continuous at
x=0 ?
Choose 1 answer:
(A) Yes
(B) No

Let $\newline$ \[ g(x)=\left\{\begin{array}{ll} \frac{1}{\cos (x)} & \text { for }-\frac{\pi}{2}

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Intermediate Value Theorem

Full solution

Q. Let

\newline

g(x)=\left\{\begin{array}{ll} \frac{1}{\cos (x)} & \text { for }-\frac{\pi}{2}<x<0 \\ \cos (x+\pi) & \text { for } x \geq 0 \end{array}\right.

\newline

g

continuous at

x=0

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B) No

Check Left-hand Limit: To determine if $g(x)$ is continuous at $x=0$ , we need to check if the left-hand limit, the right-hand limit, and the value of the function at $x=0$ are all equal.
Calculate Right-hand Limit: First, let's find the left-hand limit as $x$ approaches $0$ from the negative side ( $x \to 0^-$ ). For $x$ in the interval -\frac{\pi}{2} < x < 0, $g(x)$ is defined as $\frac{1}{\cos(x)}$ . So we need to calculate the limit of $\frac{1}{\cos(x)}$ as $x$ approaches $0$ from the left. $\newline$ $0$ $0$ .
Evaluate Function at $x=0$ : Next, we need to find the right-hand limit as $x$ approaches $0$ from the positive side ( $x \to 0^+$ ). For $x \geq 0$ , $g(x)$ is defined as $\cos(x+\pi)$ . So we need to calculate the limit of $\cos(x+\pi)$ as $x$ approaches $0$ from the right. $\newline$ $x$ $0$ .
Determine Continuity at $x=0$ : Now, we need to check the value of the function at $x=0$ . Since $x=0$ is included in the interval for the second piece of the function, $g(x)$ for $x \geq 0$ , we use the second piece to evaluate $g(0)$ . $g(0) = \cos(0+\pi) = \cos(\pi) = -1$ .
Determine Continuity at $x=0$ : Now, we need to check the value of the function at $x=0$ . Since $x=0$ is included in the interval for the second piece of the function, $g(x)$ for $x \geq 0$ , we use the second piece to evaluate $g(0)$ . $g(0) = \cos(0+\pi) = \cos(\pi) = -1$ .We have found that the left-hand limit as $x$ approaches $0$ is $1$ , the right-hand limit as $x$ approaches $0$ is $x=0$ $2$ , and the value of the function at $x=0$ is $x=0$ $2$ . Since the left-hand limit does not equal the right-hand limit, the function $g(x)$ is not continuous at $x=0$ .