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Hi, I drew this graph, but I'm not sure if it's correct, could someone check it for me, Thanks!

f'(X)= 0 at X=0 and X=2
f'(X)<0 for 0 f'(X) >0 for X<0

Draw f(X)
http://i2.photobucket.com/albums/y37/jenn_ditto/8-1.jpg

2006-11-04 13:56:24 · 8 answers · asked by Anonymous in Science & Mathematics Mathematics

8 answers

Looks like it satisfies all of the stated demands.

2006-11-04 13:59:28 · answer #1 · answered by bruinfan 7 · 0 0

(1)f'(X)= 0 at X=0 and X=2
(2)f'(X)<0 for 0 (3)f'(X) >0 for X<0

Basically 1, 2 ,3 is already get graph correctly good only the (0,0) bit misalignment.

There is another rule need to consider
f'(X) =? for X>2
if f'(X) > 0 for X>2 your graph is wrong
if f'(X) < 0 for X>2 your graph is correct

Check
rule no 2
(2)f'(X)<0 for 0 Logic consequences is
(4) f'(X)>=0 for X<=0
(5) f'(X)>=0 for X>=2

combine (4) with rule no 2 and no 1 is became no 3
(4) f'(X)>=0 for X<=0
(1)f'(X)= 0 at X=0 and X=2
(2)f'(X)<0 for 0 (3)f'(X) >0 for X<0 because f'(X)= 0 at X=0

But Combination rule (5) with rule 1 and 2
(5) f'(X)>=0 for X>=2
(1)f'(X)= 0 at X=0 and X=2
(2)f'(X)<0 for 0
will be
f'(X) > 0 for X>2 because f'(X)= 0 at X=2

Conclusion
So Your Graph is Wrong
the correct one is
f'(X) > 0 for X>2

2006-11-05 00:16:15 · answer #2 · answered by safrodin 3 · 0 0

Looks good.

Here's an equation that matches what you have:

y = -x(x-2)^2

graph it on a graphing calculator to see.

2006-11-04 22:07:45 · answer #3 · answered by modulo_function 7 · 0 0

ya it seems alright just put the arrows at the end of all 4 axes and mark X and y

2006-11-04 22:00:01 · answer #4 · answered by RichUnclePennybags 4 · 0 0

I can't tell on my screen if those say f (x) or f ' (x) (first derivative).

If it's first derivative, you are not correct. If it's just f (x) you are OK except that you slightly missed (0,0)

2006-11-04 22:02:46 · answer #5 · answered by hayharbr 7 · 1 0

it looks like the Rio Grande. you missed the whole 0,0 thing though

2006-11-04 22:35:28 · answer #6 · answered by Anonymous · 0 0

looks right, just the (0,0) part was a tad bit off

2006-11-04 22:05:09 · answer #7 · answered by Sakura_Blossoms 2 · 0 0

yes,
very good\
s

2006-11-04 22:05:35 · answer #8 · answered by locuaz 7 · 0 1

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