f (x) = x^2 +2x - 8
What is the range of f (x)?
2006-11-11 17:54:45 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
f(x) ≥ -9, or [-9, ∞)
The function has a minimum at (-1, -9), and has no upper bound.
{calculus derivation of the minimum:}
f '(x) = 2x + 2
2x + 2 = 0
x = -1 defines the minimum point
f(-1) = (-1)^2 + 2(-1) - 8
= 1-2-8
= -9
2006-11-11 17:59:06 · answer #1 · answered by Scott R 6 · 1⤊ 1⤋
There's a couple of ways to work it. The expression can be factored. You want two numbers that when multiplied equats -8 and when added equat +2. How 'bout -2, +4? These are where the expression equals zero. Thus,
f(x) = (x-2)(x+4)
The issue is the minimum. If you know calculus you can find f' = 2x+2 This is zero at x = -1, this is the minimum point. f(-1) = 1-2-8 = -9. So, the range of f is -9 to infinity.
2006-11-12 02:10:08 · answer #2 · answered by modulo_function 7 · 0⤊ 2⤋
bracket negative nine comma infinity parenthesis
2006-11-12 04:01:48 · answer #3 · answered by futuremodel21 2 · 0⤊ 1⤋