1. Determine whether this quadratic function has a maximum value or a minimum value, and then find the value:f(x) = 2x2+8x+ 5
2. Solve this inequality:(x-5)(x+2) > 0
3. True/false....The minimum value of f(x) = -x2 + 4x + 5 is f(2)
2006-09-29 16:12:35 · 7 answers · asked by KISMET 2 in Science & Mathematics ➔ Mathematics
1. It has a minimum value at x = -2
f(x) = 2x2 + 8x + 5
f'(x) = 4x + 8
0 = 4x + 8
x = -2
2. x > 5 or x < -2
3, False. The maximum value is 2.
When a parabola has a negative in front of the x2, it's upside down, so it has a maximum. When the number in front of the x2 is positive, it has a minimum.
2006-09-29 16:24:54 · answer #1 · answered by Steve A 7 · 0⤊ 0⤋
(1) y = 2*x^2 + 8*x + 5
y' = 4*x + 8 = 0 when x = -2. So (x, y) = (-2, -3) is a minimum since the coefficient of x^2 (2) is positive.
(2) (x-5)(x+2) > 0
This parabola crosses zero at -2 and 5, and since the coefficient of x^2 is 1, it must have a minimum between those two points . So for x < -2 and x > 5 the parabola is greater than 0.
(3) y = -x2 + 4x + 5
This is false; the MAXIMUM value of the function is at x = 2 since the coefficient of x^2 (-1) is negative.
2006-09-29 16:23:21 · answer #2 · answered by Joe C 3 · 0⤊ 0⤋
f(x) = 2x2 + 8x +5
f'(x) = 4x +8 = 0
=> x = -2
f(-2) = 8 -16 + 5 = -3
f(0) = 5 and so -2 must be a minima
2. x<-2 and x>5
3. F'(x) = -2x +4 = 0
=> x= 2
f(2) = -4+8+5 =9
f(0) = 5
so f(2) is a maxima not a minima
2006-09-29 16:27:17 · answer #3 · answered by vnav_in 2 · 0⤊ 0⤋
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2016-10-15 08:57:41 · answer #4 · answered by ashworth 4 · 0⤊ 0⤋
Okay the answer is totally MAYBE. God I have like NO idea! If my son were home- he is a totaly Math wiz he could tell you! The stuff he brings home might as well be German cuz I couldnt even begin to understand his math. Advanced Physics 3. He told me yesterday something about how the dog is figured in some triangle crap! Im still confused and lost! lol
2006-09-29 16:16:02 · answer #5 · answered by Anonymous · 0⤊ 2⤋
1.result:5/16;2(x2-+7x-10; false
2006-09-29 16:47:12 · answer #6 · answered by flowermieses@verizon.net 3 · 0⤊ 0⤋
True
2006-09-29 16:23:49 · answer #7 · answered by po8t1 2 · 0⤊ 1⤋