Question 140040

{{{sqrt(3-x)}}} Start with the given expression


Remember you cannot take the square root of a negative value. So that means the argument {{{3-x}}} must be greater than or equal to zero (i.e. the argument <font size=4><b>must</b></font> be positive)


{{{3-x>=0}}} Set the inner expression greater than or equal to zero


{{{-x>=0-3}}}Subtract 3 from both sides



{{{-x>=-3}}} Combine like terms on the right side



{{{x<=(-3)/(-1)}}} Divide both sides by -1 to isolate x  (note: Remember, dividing both sides by a negative number flips the inequality sign) 




{{{x<=3}}} Divide



So that means x must be less than or equal to {{{3}}} in order for x to be in the domain


So the domain in set-builder notation is

*[Tex \LARGE \textrm{\left{x|x\le3\right}}]




So here is the domain in interval notation: (-*[Tex \LARGE \infty],3]





Notice if we graph {{{y=sqrt(3-x)}}} , we get

{{{ graph( 500, 500, -10, 10, -10, 10, sqrt(3-x)) }}} notice how the graph never crosses the line {{{x=3}}}


and we can see that x must be less than or equal to {{{3}}} in order to lie on the graph. So this graphically verifies our answer.