How do you find the domain of #f(x)= 5+sqrt(9-x)#?

Answer 1

You must ensure that the argument of your square root is always bigger or equal to zero (you cannot find a real solution of a negative square root). Translated in maths means that you want: #9-x>=0# or #x<=9# that is your domain. So, you can choose 9 and numbers smaller than 9.

Graphically: graph{5+sqrt(9-x) [-28.86, 28.88, -14.43, 14.43]}

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Answer 2

Cora Alexander

To find the domain of the function ( f(x) = 5 + \sqrt{9 - x} ), we need to identify any restrictions on the values of ( x ) that would make the expression undefined. In this case, the expression under the square root, ( 9 - x ), must be non-negative (i.e., greater than or equal to zero) to ensure a real-valued result for the square root.

Set the expression under the square root greater than or equal to zero and solve for ( x ):

( 9 - x \geq 0 )

( -x \geq -9 )

Multiply both sides by -1 (and reverse the inequality sign):

( x \leq 9 )

Therefore, the domain of the function ( f(x) = 5 + \sqrt{9 - x} ) is all real numbers such that ( x \leq 9 ), or in interval notation: ( (-\infty, 9] ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.