User twilight stars ing (Suzuha) in mathematics made a note of it, to ask questions of a question, This question is
g (x) = 5 / (2 ^ x + 1) - 2, x ∈ [0, 2], y = [2 + g (x)] [1 / g (-x) - 2], y is seeking range.
I do a bit,
Simplification of the formula I obtained function is y = -25 * (2 ^ x - 1) / [(2 ^ x + 1) (3 * 2 ^ x - 2)], and the same building 29,
When 0 x =, y = 0,
When x = 2, y = -1.5,
I do not know the root of the 29th floor where 6 is coming out,
Yuzhong Hisato Hisato teacher desperately the derivative is not to determine the extreme points y and x in order to determine the [0, 2] section in the monotony?
Reply to the 34th floor Yuzhong good life Life people Teacher New Year.
This functional problem may further into y = (2 ^ x - 1) / (3 * 2 ^ 2x + 2 ^ x - 2),
If this problem is a problem of high school, you should be able to put some items in the functional disappear into a simple function, such as not fractional, but it looks as if consumers can not afford.
If the derivative to determine the extreme value, the derivative is also difficult to find out about x 0 solved when the derivative is.
So …… ？
Then on the 37th floor,
If the denominator is 2 ^ 2x - 2 * 2 ^ x + 1, it can be turned into (2 ^ x - 1) 2, then it is
y = (2 ^ x - 1) / (2 ^ x - 1) 2
= 1 / ( 2^x - 1 )
Such a knowledge can be judged in high school [0, 2] section in the monotonicity may first determine 2 ^ x - 1 monotonicity, and then determines the inverse of its monotonicity.
and many more .
37th floor of this building type and function of the coefficient of -25 less, but it does not matter, multiplied by a negative factor but so monotonous sex reversal.
Later, the landlord announced the answer, I do it again in accordance with the idea of the answer.
In this way?
y = 3 / u + 2u - 7 (1) formula
yu = 3 + 2 u 2 - 7u
2 u 2 - (7 + y) u + 3 = 0
u1 = 7 + y + [square root of [(7 + y) 2 - 24]] / 4
u2 = [7 + y - root [(7 + y) 2 - 24]] / 4
When u1 = u2, y extreme value,
7 + y + [square root of [(7 + y) 2 - 24]] / 4 = [7 + y - root [(7 + y) 2 - 24]] / 4
Root [(7 + y) 2 - 24] = - root [(7 + y) 2 - 24]
Root No. 2 * [(7 + y) 2 - 24] = 0
(7 + y) 2-24 = 0
7 + y = + (-) 2 * root (6)
y = + (-) 2 * root (6) - 7
y1 = 2 * root (6) - 7
y2 = - 2 * root (6) - 7
Y1 is substituted into the equation (1) to obtain u = root (6) / 2,
Y2 is substituted into the equation (1) to obtain u = - root (6) / 2,
Since u ∈ [1/2, 2], y1 obtained u = root (6) / 2 [1/2, 2], y2 obtained u = - root (6) / 2 is not [1 / 2, 2],
Therefore, taking y1, y1 is u ∈ [1/2, 2] in the extreme points,
Since the time when u = 1/2, y = 0, when u = 2, y = - 1.5,
Greater than 0 and -1.5 y1 = 2 * root (6) - 7, so the minimum value of y1 is u ∈ [1/2, 2] therein,
Therefore, y in u ∈ [1/2, 2] is within the range [2 * root (6) - 7, 0].
Because there are y1, y2, therefore, y in the y1, y2 are local extremum, or peak (valley).
u = y1 corresponding root (6) / 2,
y2 corresponding to u = - root (6) / 2.
It seems, quadratic or quadratic function is to determine extremes of mainstream high school ah!