How do you find domain and range for $f(x) =sqrt(4-3x) + 2$ ?

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Answer 1 · 2018-07-05T03:21:34

Domain: $(-oo,4/3]$

Range: $[2,oo)$

Let's start with the domain:

We know whatever we have under the radical cannot be negative, so we can set the following inequality:

$4-3x>=0$

Adding $3x$ to both sides gives us

$3x<=4$

$x<=4/3$ as our domain or $(-oo,4/3]$ in interval notation.

What about the range?

The lowest value that this expression can take on is $2$ , when the radical just evaluates to zero. Our range can be $2$ or greater.

$[2,oo)$ in interval notation

Hope this helps!

How do you find domain and range for f(x) =sqrt(4-3x) + 2 f(x) =sqrt(4-3x) + 2?