数学问答题!!!

junwei91

那位大大能帮我解决一些的问题?
Determine the domain and range of each function??

y=x+2
y=x+1/x+3
y=square root of x-2
y=x square

harry_lim

用addmaths以前學的functions來畫咯

笨蛋一个

原帖由 junwei91 于 25-4-2009 11:16 PM 发表
那位大大能帮我解决一些的问题?
Determine the domain and range of each function??

y=x+2
y=x+1/x+3
y=square root of x-2
y=x square

y=x+2 domain (-infinity, infinity), range (-infinity, infinity)
y=x+1/x+3 domain (-infinity, infinity), range (-infinity, infinity)
y=square root of x-2 domain [2, infinity), range [0, infinity)
y=x square domain (-infinity, infinity), range [0, infinity)

乖乖小孩

原帖由 笨蛋一个 于 26-4-2009 12:41 AM 发表


y=x+2 domain (-infinity, infinity), range (-infinity, infinity)
y=x+1/x+3 domain (-infinity, infinity), range (-infinity, infinity)
y=square root of x-2 domain [2, infinity), range [0, infinity) ...

hehe
y=x+1/x+3 domain(-infinity, infinity),range应该是[-1,1/3]才对
可是我的好像也错了！算了！找inflection point，得到minimum point就可以了！

[ 本帖最后由 乖乖小孩 于 26-4-2009 05:50 PM 编辑 ]

笨蛋一个

原帖由 乖乖小孩 于 26-4-2009 05:46 PM 发表
hehe
y=x+1/x+3 domain(-infinity, infinity),range应该是[-1,1/3]才对
可是我的好像也错了！算了！找inflection point，得到minimum point就可以了！

当lim x--> -3，range就infinity鸟

乖乖小孩

原帖由 笨蛋一个 于 26-4-2009 09:47 PM 发表


当lim x--> -3，range就infinity鸟

这样的话答案是（-infinity,1]咯

笨蛋一个

(-infinity, infinity)
lim x- -->-3=inifnity
lim x+ -->-3= - infinity

iamverynoob

y = x + 1/(x+3)

domain = (-infinity, -3), (-3, infinity)