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

Apr 15, 2015

Domain x: $\in$R, 3x$\le$4
Range y: $\in$R, y$\ge$2

The domain would be all real numbers such that 4-3x$\ge$0 Or such that 3x $\le$4, that is x$\le$ $\frac{4}{3}$. This is because the quantity under the radical sign cannot be any negative number.

For the range, solve the expression for x.
y-2= $\sqrt{4 - 3 x}$ Or,
4-3x= ${\left(y - 2\right)}^{2}$, Or
y-2= $\sqrt{4 - 3 x}$
Since 4-3x has to be $\ge 0 , y - 2 \ge$0
Hence Range would be y ;$\in$ R, y$\ge$2