Which of the following functions are continuous at x=0 ? g(x)=cot(x) h(x)=(1)/(x^(2)) Choose 1 answer: A g only (B) h only (C) Both g and h (D) Neither g nor h

Check g(x) = cot(x): To determine if g(x) = cot(x) is continuous at x=0, we need to check if the function is defined at x=0 and if the limit as x approaches 0 exists.g(x) is undefined at x=0: The cotangent function, cot(x), is the reciprocal of the tangent function, tan(x). Since tan(0) = 0, cot(x) = 1/tan(x) is undefined at x=0 because we cannot divide by zero.Check h(x) = 1/x^2: Since g(x) = cot(x) is undefined at x=0, g(x) is not continuous at x=0.h(x) is undefined at x=0: Now, let's check if h(x) = 1/x^2 is continuous at x=0. We need to see if the function is defined at x=0 and if the limit as x approaches 0 exists.Conclusion: The function h(x) = 1/x^2 is undefined at x=0 because we cannot divide by zero when x=0.Since h(x) = 1/x^2 is undefined at x=0, h(x) is not continuous at x=0.Both functions g(x) = cot(x) and h(x) = 1/x^2 are not continuous at x=0 because they are both undefined at that point.

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Which of the following functions are continuous at
x=0 ?

g(x)=cot(x)

h(x)=(1)/(x^(2))
Choose 1 answer:
A
g only
(B)
h only
(C) Both
g and
h
(D) Neither
g nor
h

Which of the following functions are continuous at $x=0$ ? $\newline$ $g(x)=\cot (x)$ $\newline$ $h(x)=\frac{1}{x^{2}}$ $\newline$ Choose $1$ answer: $\newline$ (A) $g$ only $\newline$ (B) $h$ only $\newline$ (C) Both $g$ and $h$ $\newline$ (D) Neither $g$ nor $h$

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Math Problems

Algebra 1

Domain and range of quadratic functions: equations

Full solution

Q. Which of the following functions are continuous at

x=0

\newline

g(x)=\cot (x)

\newline

h(x)=\frac{1}{x^{2}}

\newline

Choose

1

answer:

\newline

(A)

g

only

\newline

(B)

h

only

\newline

g

and

h

\newline

(D) Neither

g

nor

h

Check $g(x) = \cot(x)$ : To determine if $g(x) = \cot(x)$ is continuous at $x=0$ , we need to check if the function is defined at $x=0$ and if the limit as $x$ approaches $0$ exists.
$g(x)$ is undefined at $x=0$ : The cotangent function, $\text{cot}(x)$ , is the reciprocal of the tangent function, $\text{tan}(x)$ . Since $\text{tan}(0) = 0$ , $\text{cot}(x) = \frac{1}{\text{tan}(x)}$ is undefined at $x=0$ because we cannot divide by zero.
Check $h(x) = \frac{1}{x^2}$ : Since $g(x) = \cot(x)$ is undefined at $x=0$ , $g(x)$ is not continuous at $x=0$ .
ext{ extit{h}}(x) is undefined at ext{ extit{x}}= $0$ : Now, let's check if ext{ extit{h}}(x) = rac{ $1$ }{x^ $2$ } is continuous at ext{ extit{x}}= $0$ . We need to see if the function is defined at ext{ extit{x}}= $0$ and if the limit as ext{ extit{x}} approaches $0$ exists.
Conclusion: The function $h(x) = \frac{1}{x^2}$ is undefined at $x=0$ because we cannot divide by zero when $x=0$ .
Conclusion: The function $h(x) = \frac{1}{x^2}$ is undefined at $x=0$ because we cannot divide by zero when $x=0$ .Since $h(x) = \frac{1}{x^2}$ is undefined at $x=0$ , $h(x)$ is not continuous at $x=0$ .
Conclusion: The function $h(x) = \frac{1}{x^2}$ is undefined at $x=0$ because we cannot divide by zero when $x=0$ .Since $h(x) = \frac{1}{x^2}$ is undefined at $x=0$ , $h(x)$ is not continuous at $x=0$ .Both functions $g(x) = \cot(x)$ and $h(x) = \frac{1}{x^2}$ are not continuous at $x=0$ because they are both undefined at that point.