Find the Domain and Range f(x)=(3x^2+1)/(x^2+x+9)

Problem

ƒ(x)=(3*x^2+1)/(x^2+x+9)

Solution

1. Identify the domain by finding where the denominator x^2+x+9 is equal to zero.

2. Calculate the discriminant of the denominator D=b^2−4*a*c=1^2−4*(1)*(9)=−35

3. Determine that since the discriminant is negative, the denominator has no real roots and is always positive, meaning the domain is all real numbers.

4. Express the domain in interval notation as (−∞,∞)

5. Find the range by setting y=(3*x^2+1)/(x^2+x+9) and rearranging into a quadratic equation in terms of x (y−3)*x^2+y*x+(9*y−1)=0

6. Apply the condition for real x by requiring the discriminant of this quadratic to be non-negative: y^2−4*(y−3)*(9*y−1)≥0

7. Simplify the inequality to −35*y^2+112*y−12≥0 which factors as −(7*y−2)*(5*y−6)≥0

8. Solve the inequality to find the range: 2/7≤y≤6/5

Final Answer

Domain: *(−∞,∞), Range: *[2/7,6/5]

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