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$Let f(x)={[(2)/(x^(2))," for "x <= -1],[(x+3)/(cos(x+1))," for "-1 < x < (pi-2)/(2)]:} Is f continuous at x=-1 ? Choose 1 answer: (A) Yes (B) No$

Let $\newline$ \( f(x)=\left\{\begin{array}{ll}\frac{2}{x^{2}} & \text { for } x \leq-1 \\ \frac{x+3}{\cos (x+1)} & \text { for }-1

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Calculus

Intermediate Value Theorem

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Q. Let

\newline

f(x)=\left\{\begin{array}{ll}\frac{2}{x^{2}} & \text { for } x \leq-1 \\ \frac{x+3}{\cos (x+1)} & \text { for }-1<x<\frac{\pi-2}{2}\end{array}\right.

\newline

f

continuous at

x=-1

\newline

Choose

1

answer:

\newline

(A) Yes

\newline

(B)

\mathrm{No}

Conditions for Continuity: To determine if the function $f(x)$ is continuous at $x=-1$ , we need to check if the following three conditions are met: $\newline$ $1$ . $f(-1)$ is defined. $\newline$ $2$ . The limit of $f(x)$ as $x$ approaches $-1$ from the left exists. $\newline$ $3$ . The limit of $f(x)$ as $x$ approaches $-1$ from the right exists and is equal to the limit from the left and to $f(-1)$ .
Finding $f(-1)$ : First, we will find $f(-1)$ using the definition of the function for $x \leq -1$ , which is $f(x) = \frac{2}{x^2}$ . $\newline$ Substitute $x = -1$ into the function to get $f(-1) = \frac{2}{(-1)^2} = \frac{2}{1} = 2$ .
Limit from the Left: Next, we need to find the limit of $f(x)$ as $x$ approaches $-1$ from the left. Since the function for $x \leq -1$ is $f(x) = \frac{2}{x^2}$ , the limit as $x$ approaches $-1$ from the left is the same as $f(-1)$ , which is $2$ .
Limit from the Right: Now, we need to find the limit of $f(x)$ as $x$ approaches $-1$ from the right. For -1 < x < (\pi-2)/2, the function is defined as $f(x) = (x+3)/\cos(x+1)$ . $\newline$ We need to evaluate the limit of $(x+3)/\cos(x+1)$ as $x$ approaches $-1$ from the right.
Conclusion: As $x$ approaches $-1$ from the right, the numerator $(x+3)$ approaches $2$ . The denominator $\cos(x+1)$ approaches $\cos(0)$ , which is $1$ . Therefore, the limit of $f(x)$ as $x$ approaches $-1$ from the right is $-1$ $0$ .
Conclusion: As $x$ approaches $-1$ from the right, the numerator $(x+3)$ approaches $2$ . The denominator $\cos(x+1)$ approaches $\cos(0)$ , which is $1$ . Therefore, the limit of $f(x)$ as $x$ approaches $-1$ from the right is $-1$ $0$ .Since the limit from the left is $2$ , the limit from the right is $2$ , and $-1$ $3$ is also $2$ , all three conditions for continuity are satisfied. Therefore, $f(x)$ is continuous at $-1$ $6$ .