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

Check left-hand limit: To determine if g(x) is continuous at x=-2, we need to check if the left-hand limit, the right-hand limit, and the value of the function at x=-2 are all equal.Check right-hand limit: First, we find the left-hand limit as x approaches -2 from the left. Since for x -2, g(x) = -(cos(x+2))/2. We substitute x with a value slightly greater than -2, such as -1.99, and use the fact that cos(0) = 1 to see that as x approaches -2, g(x) approaches -1/2.Conclusion: Now, we need to find the value of the function at x=-2. Since x=-2 falls in the first case of the piecewise function, we use g(x) = 1/x and find that g(-2) = 1/(-2) = -1/2.Since the left-hand limit, the right-hand limit, and the value of the function at x=-2 are all equal to -1/2, the function g(x) is continuous at x=-2.

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Let

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

Let $\newline$ $g(x)=\left\{\begin{array}{ll} \frac{1}{x} & \text { for } x \leq-2 \\ -\frac{\cos (x+2)}{2} & \text { for } x>-2 \end{array}\right.$ $\newline$ Is $g$ continuous at $x=-2$ ? $\newline$ Choose $1$ answer: $\newline$ (A) Yes $\newline$ (B) No

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Math Problems

Calculus

Intermediate Value Theorem

Full solution

Q. Let

\newline

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

\newline

g

continuous at

x=-2

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B) No

Check left-hand limit: To determine if $g(x)$ is continuous at $x=-2$ , we need to check if the left-hand limit, the right-hand limit, and the value of the function at $x=-2$ are all equal.
Check right-hand limit: First, we find the left-hand limit as $x$ approaches $-2$ from the left. Since for $x \leq -2$ , $g(x) = \frac{1}{x}$ , we substitute $x$ with a value slightly less than $-2$ , such as $-2.01$ , and see that as $x$ approaches $-2$ , $g(x)$ approaches $-2$ $0$ .
Find value at $x=-2$ : Next, we find the right-hand limit as $x$ approaches $-2$ from the right. For x > -2, $g(x) = -\frac{\cos(x+2)}{2}$ . We substitute $x$ with a value slightly greater than $-2$ , such as $-1.99$ , and use the fact that $\cos(0) = 1$ to see that as $x$ approaches $-2$ , $x$ $1$ approaches $x$ $2$ .
Conclusion: Now, we need to find the value of the function at $x=-2$ . Since $x=-2$ falls in the first case of the piecewise function, we use $g(x) = \frac{1}{x}$ and find that $g(-2) = \frac{1}{(-2)} = -\frac{1}{2}$ .
Conclusion: Now, we need to find the value of the function at $x=-2$ . Since $x=-2$ falls in the first case of the piecewise function, we use $g(x) = \frac{1}{x}$ and find that $g(-2) = \frac{1}{(-2)} = -\frac{1}{2}$ .Since the left-hand limit, the right-hand limit, and the value of the function at $x=-2$ are all equal to $-\frac{1}{2}$ , the function $g(x)$ is continuous at $x=-2$ .