What is the domain and range of #f(x) = sqrt(4-3x) + 2#?

1 Answer
bp
Apr 15, 2015

Domain x: #in#R, 3x#<=#4
Range y: #in#R, y#>=#2

The domain would be all real numbers such that 4-3x#>=#0 Or such that 3x #<=#4, that is x#<=# #4/3#. This is because the quantity under the radical sign cannot be any negative number.

For the range, solve the expression for x.
y-2= #sqrt(4-3x)# Or,
4-3x= #(y-2)^2#, Or
y-2= #sqrt(4-3x)#
Since 4-3x has to be #>=0, y-2 >=#0
Hence Range would be y ;#in# R, y#>=#2

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