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Small Calculus Question.

Blogs > clazziquai
Post a Reply
clazziquai
Profile Blog Joined October 2007
6685 Posts
October 01 2008 03:43 GMT
#1
Problem 1. a) Suppose that f(x) = x^2 and g(x) = 2^x. Compute f(−2), g(−2), f(5), and
g(5). According to the Intermediate Value Theorem and the function values computed,
what is the smallest number of roots the equation f(x) = g(x) can have?

b) Suppose still that f(x) = x^2 and g(x) = 2^x. Graph y = f(x) and y = g(x) carefully on
the interval −2 <= x <= 5. How many roots does the equation f(x) = g(x) appear to have?

For Part A I understand everything and I believe that the smallest number of roots would be 1 right? That's what my friend and I concluded but can anyone verify this?

And for Part B, would there be two roots? Again, can you guys verify my answers? (: Thank you very much.

#1 Sea.Really Fan / #1 Nesh Fan / Terran Forever~
micronesia
Profile Blog Joined July 2006
United States24677 Posts
Last Edited: 2008-10-01 03:51:57
October 01 2008 03:48 GMT
#2
Whenever a function becomes higher than another, there has to be at least one intersection...

edit: I plugged them in and it seems f is higher at -2 and g is higher at 5, but there are 3 intersections instead of one (had to be an odd number since they are both continuous)
ModeratorThere are animal crackers for people and there are people crackers for animals.
infinity21 *
Profile Blog Joined October 2006
Canada6683 Posts
Last Edited: 2008-10-01 04:29:01
October 01 2008 04:28 GMT
#3
well, f(-2) = 4 and g(-2) = 1/4 so f > g @ -2
f(5) = 25 and g(5) = 32 so f < g @ 5
So using the IVT, you are guaranteed at least 1 root

If you graph the two functions, you'll see that there are 3 intersections
http://img210.imageshack.us/my.php?image=graphpe0.jpg

Official Entusman #21
infinity21 *
Profile Blog Joined October 2006
Canada6683 Posts
October 01 2008 04:30 GMT
#4
one of the roots is between -1 and 0, the other two are at 2 and 4
Official Entusman #21
vAltyR
Profile Blog Joined July 2008
United States581 Posts
October 01 2008 04:32 GMT
#5
for part A, the Intermediate Value Theorem says they must intersect at least once. It actually is three times, but according to the IVT and the values computed, the smallest number of roots is one.

for part B, the answer is three roots, according to micronesia. *pulls out grapher and double checks* yeah, it's three. use a graphing calculator to graph both equations and trace along one until you find all three intersections.
내 호버크라프트는 장어로 가득 차 있어요
doghunter
Profile Blog Joined July 2008
United States23 Posts
October 01 2008 04:36 GMT
#6
For part B, you can use Descartes' rule of signs.

Or you can look at 2^x=x^2 and be like, oh if x=0 there is one solution and if x!=0 then c=x^2 so there are two solutions
clazziquai
Profile Blog Joined October 2007
6685 Posts
Last Edited: 2008-10-01 04:43:57
October 01 2008 04:43 GMT
#7
ty guys~
#1 Sea.Really Fan / #1 Nesh Fan / Terran Forever~
infinity21 *
Profile Blog Joined October 2006
Canada6683 Posts
October 01 2008 04:55 GMT
#8
lol he asks for calc help and then requests ban. What if he has more questions?
Official Entusman #21
Klockan3
Profile Blog Joined July 2007
Sweden2866 Posts
October 01 2008 05:22 GMT
#9
On October 01 2008 13:55 infinity21 wrote:
lol he asks for calc help and then requests ban. What if he has more questions?

Do people actually request bans?
infinity21 *
Profile Blog Joined October 2006
Canada6683 Posts
October 01 2008 05:24 GMT
#10
On October 01 2008 13:46 TL.net Bot wrote:
clazziquai was just Temp banned for 1 week by SonuvBob.

That account was created on 2007-10-07 13:34:57 and had 2510 posts.

Reason: Request.
Official Entusman #21
Not_Computer
Profile Blog Joined January 2007
Canada2277 Posts
October 01 2008 05:26 GMT
#11
i guess he wanted to concentrate on his studies and stay away from TL?

o.o;
"Jaedong hyung better be ready. I'm going to order the most expensive dinner in Korea."
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