Question

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

Answer 1

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