Find domain of f(x)=1/√x^2+1

So I put a lot of effort into this answer.

Step-by-step explanation:

√X^2–1 is not equal to 0 , X^2–1 is not equal to 0 also, (X+1)(X-1) is not equal to 0 & X is not equal to -1 & 1 .

For root function to be defined…….

X^2–1>=0…(1)

(X+1)(X-1)>=0...(2)

i.e. X lies between (-infinite, -1)U(1,infinite)

Domain (f) ={X|X€R, X€(-infinite, -1)U(1,infinite)}

Rewriting function,

Y=1/√X^2–1

Y^2=1/X^2–1

X^2–1=1/Y^2

X^2=(Y^2+1)/Y^2

X=√Y^2+1/Y

for this to be defined Y can't be equal to 0 and for all values of Y, √Y^2+1>0

Range (f)={Y|Y€R}=R-{0}……

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58

dont know

step-by-step explanation: