如图,A,B,C为函数y=log1/2 x的图像上的三点,它们的横坐标分别是t,t+2,t+4(t≥1)如图,在函数y=log1/2 X的图向上有A,B,C三点,它们的横坐标分别为t,t+2,t+4(t≥0) 1.若三角形ABC的面积为S,求S=f(t) 2.
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![如图,A,B,C为函数y=log1/2 x的图像上的三点,它们的横坐标分别是t,t+2,t+4(t≥1)如图,在函数y=log1/2 X的图向上有A,B,C三点,它们的横坐标分别为t,t+2,t+4(t≥0) 1.若三角形ABC的面积为S,求S=f(t) 2.](/uploads/image/z/3392907-51-7.jpg?t=%E5%A6%82%E5%9B%BE%2CA%2CB%2CC%E4%B8%BA%E5%87%BD%E6%95%B0y%3Dlog1%2F2+x%E7%9A%84%E5%9B%BE%E5%83%8F%E4%B8%8A%E7%9A%84%E4%B8%89%E7%82%B9%2C%E5%AE%83%E4%BB%AC%E7%9A%84%E6%A8%AA%E5%9D%90%E6%A0%87%E5%88%86%E5%88%AB%E6%98%AFt%2Ct%2B2%2Ct%2B4%28t%E2%89%A51%EF%BC%89%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E5%87%BD%E6%95%B0y%3Dlog1%2F2+X%E7%9A%84%E5%9B%BE%E5%90%91%E4%B8%8A%E6%9C%89A%2CB%2CC%E4%B8%89%E7%82%B9%2C%E5%AE%83%E4%BB%AC%E7%9A%84%E6%A8%AA%E5%9D%90%E6%A0%87%E5%88%86%E5%88%AB%E4%B8%BAt%2Ct%2B2%2Ct%2B4%EF%BC%88t%E2%89%A50%EF%BC%89+1.%E8%8B%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E9%9D%A2%E7%A7%AF%E4%B8%BAS%2C%E6%B1%82S%EF%BC%9Df%28t%EF%BC%89+2%EF%BC%8E)
如图,A,B,C为函数y=log1/2 x的图像上的三点,它们的横坐标分别是t,t+2,t+4(t≥1)如图,在函数y=log1/2 X的图向上有A,B,C三点,它们的横坐标分别为t,t+2,t+4(t≥0) 1.若三角形ABC的面积为S,求S=f(t) 2.
如图,A,B,C为函数y=log1/2 x的图像上的三点,它们的横坐标分别是t,t+2,t+4(t≥1)
如图,在函数y=log1/2 X的图向上有A,B,C三点,它们的横坐标分别为t,t+2,t+4(t≥0)
1.若三角形ABC的面积为S,求S=f(t)
2.判断S=f(t)的单调性
3.求S=f(t)的最大值
第一题要十分详细,因为我没学过,
如图,A,B,C为函数y=log1/2 x的图像上的三点,它们的横坐标分别是t,t+2,t+4(t≥1)如图,在函数y=log1/2 X的图向上有A,B,C三点,它们的横坐标分别为t,t+2,t+4(t≥0) 1.若三角形ABC的面积为S,求S=f(t) 2.
(1)A(t,log(1/2)(t)),B(t+2,log(1/2)(t+2)),C(t+4,log(1/2)(t+4)),
当00) 在(0,+∞)上为减函数.
(3)无最大值.楼主,怕是t≥1吧?(当t=1时,S大=log(1/2)(5/9) .)
要积分才能解啊。LZ.你是哪个知识程度?
(1)A(t,log(1/2)(t)),B(t+2,log(1/2)(t+2)),C(t+4,log(1/2)(t+4)),
当0
=|log(1/2)(t+2)-log(1/2)(t)|+(|log(1/2)(t+2)-log(1...
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(1)A(t,log(1/2)(t)),B(t+2,log(1/2)(t+2)),C(t+4,log(1/2)(t+4)),
当0
=|log(1/2)(t+2)-log(1/2)(t)|+(|log(1/2)(t+2)-log(1/2)(t)|+|log(1/2)(t+4)-log(1/2)(t)|)-2|log(1/2)(t+4)-log(1/2)(t)|
=log(1/2)[t(t+4)/(t+2)^2
=log(1/2)[1-4/(t+2)^2]
当t≥1时
S=S梯形ABB1A1+S梯形BCC1B1-S梯形ACC1A1
=(|log(1/2)t|+|log(1/2)(t+2)|)+(|log(1/2)(t+2)|+|log(1/2)(t+4)|)-2(|log(1/2)t|+|log(1/2)(t+4)|),
S=log(1/2)[t(t+4)/(t+2)^2
=log(1/2)[1-4/(t+2)^2]
综上,S=log(1/2)[1-4/(t+2)^2] (t>0)
(2)S=log(1/2)[1-4/(t+2)^2] (t>0) 在(0,+∞)上为减函数。
(3)无最大值. 楼主,怕是t≥1吧?(当t=1时,S大=log(1/2)(5/9) 。)
收起