Math  /  Algebra

QuestionConsider a function f(x)=3x+a+2f(x)=-\sqrt{3x+a+2}.
c) In the space below, determine an expression representing the domain and range of f(x)f(x).
Do not provide the full form, provide the inequality portions only! For example, your domain could be x<5x<5 and your range could be y75y \geq-75.
D: \square
R: \square
d) In the space below, determine an expression representing the xx-intercept of f(x)f(x). state the xx-coordinate only. x=x= \square

Studdy Solution
To find the x x -intercept, set f(x)=0 f(x) = 0 and solve for x x : 0=3x+a+2 0 = -\sqrt{3x + a + 2} This implies 3x+a+2=0 \sqrt{3x + a + 2} = 0 .
The domain inequality is: 3x+a+20 3x + a + 2 \geq 0
The range inequality is: y0 y \leq 0
The x x -intercept is found by solving: 3x+a+2=0 3x + a + 2 = 0

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