Question

is continuous at x = 0.
q7(*):
describe all points for which the function is continuous: ( f(x) = \frac{1}{x^2 - 1} ).
q8(*):
describe all points for which this function is continuous: ( f(x) = \frac{1}{sqrt{x^2 - 1}} ).
q9(*):
describe all points for which this function is continuous: ( \frac{1}{sqrt{1 + cos(x)}} ).

Explanation:

Q[7]

Step1: Find where denominator is zero

A rational function \( f(x)=\frac{1}{x^2 - 1} \) is undefined when denominator \( x^2-1 = 0 \). Solve \( x^2-1=0 \), we get \( x^2=1 \), so \( x=\pm1 \).

Step2: Determine continuity

A rational function is continuous everywhere except where it is undefined. So the function is continuous for all real numbers \( x \) such that \( x
eq - 1 \) and \( x
eq1 \).

Step1: Analyze the domain restrictions

For the function \( f(x)=\frac{1}{\sqrt{x^2 - 1}} \), we have two restrictions: the expression under the square root must be positive (since it's in the denominator, it can't be zero either) and the denominator can't be zero. So we need \( x^2-1>0 \).

Step2: Solve the inequality

Solve \( x^2 - 1>0 \), factor it as \( (x - 1)(x + 1)>0 \). The critical points are \( x=-1 \) and \( x = 1 \). Testing intervals:

• For \( x<-1 \), say \( x=-2 \), \( (-2 - 1)(-2 + 1)=(-3)\times(-1)=3>0 \).
• For \( - 1
• For \( x>1 \), say \( x = 2 \), \( (2 - 1)(2 + 1)=(1)\times(3)=3>0 \).

So \( x^2-1>0 \) when \( x<-1 \) or \( x>1 \).

Step3: Determine continuity

The function is continuous on its domain. So the function is continuous for all real numbers \( x \) such that \( x<-1 \) or \( x>1 \), i.e., the intervals \( (-\infty,-1)\cup(1,+\infty) \).

Step1: Analyze the domain restrictions

For the function \( f(x)=\frac{1}{\sqrt{1+\cos(x)}} \), we need two things: the expression under the square root \( 1+\cos(x)\geq0 \) (since it's in a square root) and the denominator \( \sqrt{1 + \cos(x)}
eq0 \), so \( 1+\cos(x)>0 \).

Step2: Solve the inequality \( 1+\cos(x)>0 \)

We know that \( \cos(x)\in[-1,1] \) for all real \( x \). \( 1+\cos(x)>0 \) is equivalent to \( \cos(x)>- 1 \). The equation \( \cos(x)=-1 \) when \( x=(2n + 1)\pi \), where \( n\in\mathbb{Z} \) (integers). So \( \cos(x)>-1 \) for all real \( x \) except \( x=(2n + 1)\pi,n\in\mathbb{Z} \).

Step3: Determine continuity

The function is continuous on its domain. So the function is continuous for all real numbers \( x \) except \( x=(2n + 1)\pi \), where \( n\in\mathbb{Z} \).

Answer:

The function \( f(x)=\frac{1}{x^2 - 1} \) is continuous at all real numbers \( x \) with \( x
eq - 1 \) and \( x
eq1 \), i.e., the intervals \( (-\infty,-1)\cup(-1,1)\cup(1,+\infty) \).

Q[8]