seriously... its like every problem is different! :[ Now i need to use the second derivative test to find the relative maxma and minima...
y=-2x^2 + 8x + 25
okay. so i can find the first and second derivatives of these just fine.. but the second derivative turns into a constant so how do i go about using that to figure out the extremas?
2007-11-27 05:39:51 · 4 answers · asked by chelsie r 2 in Science & Mathematics ➔ Mathematics
-4x + 8 = f'(x)
f''(x) = -4
Since this is a quadratic equation, you know that the only critical value is when the first derivative is equal to zero at the max or min of the parabola.
You know that occurs when x = 2.
When you take the second derivative, it's always less than zero, so the curve is always concave down.
It's an upside down parabola.
2007-11-27 05:49:11 · answer #1 · answered by Useless Knowledge Goddess 4 · 2⤊ 2⤋
So you found that y'' = -4.
the Fact that it is a negative tells you that the curveature of the function is "concave down" So the critical point, where y'=0, in this case -4x + 8=0, is a MAXIMUM
You don't need the second derivative y'' to be a changing valued function. The fact that it is a constant just means that the function always curves the same way. In your case you have a parabola that curves down so it's vertex is it's highest point.
2007-11-27 05:53:06 · answer #2 · answered by Anonymous · 1⤊ 2⤋
f `(x) = - 4x + 8
f "(x) = - 4
- 4x + 8 = 0 for turning point
x = 2 for turning point.
f "(2) = - 4 (gives a Max. turning point)
f (2) = - 8 + 16 + 25 = 33
(2,33) is a maximum turning point.
In answer to what you say, it is the fact that f "(x) is -ve that dictates that (2,33) is a MAXIMUM turning point.
2007-11-27 05:55:33 · answer #3 · answered by Como 7 · 3⤊ 1⤋
y'' = -4, so the critical point you find by setting y' = 0 would be a relative maximum.
2007-11-27 05:48:26 · answer #4 · answered by ben e 7 · 1⤊ 2⤋