請教[不等式]一題

[ching]

f(x)=2^(x+2) – 3 * 4^x ， -1 <= x <= 0，則f(x)的最大值?(ans: 4/3)
謝謝

[thepiano]

f(x) = 2^(x + 2) - 3 * 4^x = 4 * 2^x - 3 * (2^x)^2

令 t = 2^x，1/2 ≦ t ≦ 1

f(x) 改寫成 f(t) = 4t - 3t^2 = -3t^2 + 4t

t = - 4 / (-6) = 2/3 時，f(t) 有最大值 4/3

[ching]

謝謝鋼琴老師
我想請問
t = - 4 / (-6) = 2/3 時
是怎麼算的?
我每次都是f(t)= -3t+4t = (-3){t^2 - (4/3)t + (2/3)^2} + (4/3) = (-3)[t-(2/3)]^2 + (4/3)
您的方法好像快多了,不知 t = - 4 / (-6) 如呵解?
不好意思,我的數學不好,麻煩您了!
謝謝

[thepiano]

f(x) = ax^2 + bx + c　(a≠0)

= a[x^2 + (b/a)x] + c

= a[x^2 + (b/a)x + (b/2a)^2] + c - a * (b/2a)^2

= a[x + (b/2a)]^2 + [-(b^2 - 4ac)/(4a)]

(1) a ＞ 0
x = -b/2a 時，f(x) 有極小值 -(b^2 - 4ac)/(4a)

(2) a ＜ 0
x = -b/2a 時，f(x) 有極大值 -(b^2 - 4ac)/(4a)

[ching]

原來如此,真是謝謝您!