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Math  /  Algebra

QuestionDetermine the real values of xx that cause the function to be (a) zero, (b) undefined, (c) positive, f(x)=xx+12f(x)=x \sqrt{x+12} A. x=−12x=-12 B. x<12x<12 C. x<−12x<-12 D. x=12x=12 (c) What value(s) of x cause the function to be positive? Choose the correct answer below. A. (∞,−12)∪(12,∞)(\infty,-12) \cup(12, \infty) B. (−12,∞)(-12, \infty) C. (12,∞)(12, \infty) D. (0,∞)(0, \infty)

Studdy Solution
To find when the function is positive, determine when both x>0 x > 0 and x+12≥0 x + 12 \geq 0 :
1. x>0 x > 0 ensures x x is positive.
2. x+12≥0 x + 12 \geq 0 simplifies to x≥−12 x \geq -12 .
The intersection of these conditions is x>0 x > 0 .
The correct answer for when the function is positive is:
D. (0,∞)(0, \infty)

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