It is known that the maximum value of the function f (x) = a ^ x (a > 0, and a ≠ 1) in the interval [1,2] is m, and the minimum value is n 1, if M + n = 6, find the value of real number a If M = 2n, find the value of real number a Wait online

Discussion on Classification: for the base a, 0

If the maximum value of function f (x) = - x ^ 3 + 12x + A in the interval [- 1,1] is 2, then its minimum value in the interval is 2

f(x)=-x^3+12x+a
f'(x)=-3x²+12=0
-3(x+2)(x-2)=0
X = - 2 or x = 2
When x ∈ [- 1,1],
f'(x)>0
therefore
The function is an increasing function, that is, the maximum value = f (1) = - 1 + 12 + a = 2
a=-9
therefore
Minimum value = f (- 1) = 1-12 + a = 1-12-9 = - 20

Find the maximum and minimum values of the function f (x) = x ^ 3-12x + 2 in the interval [- 3. Three thirds]

The derivation 3x ^ 2-12 = 0, in the interval [- 3,3 / 2], x = - 2
F (x) is a monotone function in [- 3, - 2], [- 2,3 / 2]
f(-2)=18
f(-3)=11
f(3/2)=-101/8
therefore
The maximum value is f (- 2) = 18
The minimum value is f (3 / 2) = - 101 / 8

The detailed process is needed: 1) function f (x) = 6-12x + X & # 179; what is the maximum value and the minimum value in the interval [- 3,1]? 2
② It is known that F1F2 is two focal points of ellipse X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1 (a ＞ B ＞ 0). If there is a point P on the ellipse and the angle f1pf2 = 120 °, what is the range of eccentricity of the ellipse, Find linear l equation. Detailed and timely process can be appropriate to add reward

1. The derivative of FX is 3x ^ 2-12,
Make it 0
X = plus or minus 2
When x is less than - 2, the derivative is greater than 0 and increases in the interval
-2

It is known that the maximum value of the function f (x) = a ^ x (a > 0, and a ≠ 1) in the interval [1,2] is m, and the minimum value is n 1, if M + n = 6, find the value of real number a If M = 2n, find the value of real number a Wait online