find all values of c such that f(x) = (x+1)/(x^2+2cx+4) has domain R
and what domain is R?
it just says domain R. I guess all real numbers?
To find all values of c such that the function f(x) = (x+1)/(x^2+2cx+4) has a domain of R (all real numbers), we need to consider the values of x for which the denominator of the function is non-zero.
The denominator (x^2 + 2cx + 4) can be factored as a quadratic expression. We can use the quadratic formula to find the roots of this quadratic equation:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = 2c, and c = 4. Substituting these values into the quadratic formula, we get:
x = (-2c ± √((2c)^2 - 4(1)(4))) / (2(1))
x = (-2c ± √(4c^2 - 16)) / 2
x = -c ± √(c^2 - 4)
For the function f(x) to have a real domain, this expression inside the square root (c^2 - 4) must be greater than or equal to zero. In other words:
c^2 - 4 ≥ 0
We can solve this inequality by factoring:
(c - 2)(c + 2) ≥ 0
To determine the values of c that satisfy this inequality, we consider the sign of the expression (c - 2)(c + 2) for different ranges of c.
1. When c < -2: Both factors (c - 2) and (c + 2) are negative, so their product is positive. This range of c satisfies the inequality.
2. When -2 ≤ c ≤ 2: The factor (c - 2) becomes non-positive, while the factor (c + 2) remains positive. Thus, this range of c does not satisfy the inequality.
3. When c > 2: Both factors (c - 2) and (c + 2) are positive, so their product is positive. This range of c also satisfies the inequality.
Therefore, the values of c that make the function f(x) = (x+1)/(x^2+2cx+4) have a real domain are c < -2 or c > 2.