DOMAIN AND RANGE

What is the range of the real function f defined by f(x) =√(x-1)?     

Given Y=√(x-1)

Range is the value of y for the function to be defined.

convert x is in the form of y

i.e.

y²=x-1

or x=1+y²

hence the x is always defined for all real value of y

so the range of function is y∈R i.e. all real value

What is the domain function f, where f(x) = √(1-x) (x-2)?

Given y=√(1-x) (x-2)

for function to be defined i.e. y to defined domain is value of x for which y is defined

y is defined when value under squre root is greater than and equal to zero

i.e.

(1-x)(x-2)≥0

or (x-1)(x-2)≤0

or 1≤x≤2

What will be the range of the function f(x) =1/ (√ (x^2-1))?

Range is the value of the y for which x is defined

rewritting as y=1/ (√ (x^2-1))

or y²=1/(x^2-1)

or x²-1=1/y²

or x²=1/y²+1

or x=√((1/y²)+1

hence the x is defined for all value of y except when y is zero

range y∈R-(1)

What are the domain and range of the function y=-(x-2)^2 +2?

Domain is the value of x for which function is defined, here for all value of x i.e. all real value of x.

Range is the value of y for which function is defined.

(y-2)=-(x-2)²

or (2-y)=(x-2)²

or (x-2)=√2-y

for this function to defined (2-y)≥0

or (y-2)≤0

or y≤2

What is the range of f(x)=(sinx)sinx+1?

range is the value of y for which function to be defined 

min value of sinx=0 when x=0

hence min value when x=0 is 0⁰+1

which is not defined hence solve by limit

limx→0 (sinx)^sinx 

^{or taking log on both sides}

^{lny=limx→0ln}(sinx)^sinx 

^{or lny=}limx→0sinxln(sinx)

or lny=limx→0ln(sinx)/coescx

solve using L hospital rule

lny=limx→0cosx/(-cosecx.cotx)(sinx)

lny=-1

or y=1/e

hence min value of y is (1/e)Λ(1/e)+1

and max value when x=90°

ie. y(max)=1+1=2

range is ((1/e)Λ(1/e)+1),2)