Question

find the domain and range of f(x)=1+x(square)

Answer 1

\[f(x)=1+x^2\]
as there is no x for which f(x) is undefined
domain= all real numbers
the minimum value of f(x)= 1 when x=0
so range = [1, \(\infty\)) or f(x)>=1

Answer 2

domain: all reals
range : [1 , oo)

Answer 3

ash, good explanation ;)

Answer 4

we take as the domain the largest possible set of real numbers. since there are no restrictions (no points that are undefined), the domain is all reals

Answer 5

, it is assumed that the domain is the largest possible set of reals

Answer 6

izzati you got it?

Answer 7

actually i dont understand...it is in exam we should write it like that??

Answer 8

yeah izzati

Answer 9

how you got the range..explain plez..

Answer 10

We have
\[f(x)=1+x^2\]
we know that x^2 's minimum value is 0 whether x>0 or x<0 , value of square of x will be greater than 0
so its minimum value is 0
so f(x)'s minimum value is 1+0=1
when x^2 increases, f(x) also increases
when x -----> infinity , f(x) ----> infinity
no maximum limit of f(x)
so
f(x) 's range is greater than or equal to 1
f(x)>=1
or
range is [1, \(\infty\))

Answer 11

ok.i got it thanx