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Can someone help me with these calc problems. I'm sooo lost. Please don't insult me if you think these are easy. I know I'm not very good at this. How do you....

a.)
simplify (x^3-125)/(x-5) to remove the discontinuity.

b.)
(same formula as in a) Use appropriate limit properties to find the limit of m(x) as x approaches 5.

also,

c.) Using the intermediate value theorem show that the equation x=cosx has at least one solution in the interval [0,Pi/2]

and...

d.) f(x)=x^4 Evaluate q(2) and q(3) and use the intermediate value theorem to show there is a # between 2 and 3 that is exactly equal to the fourth root of 51.

I want to learn this stuff, but apparantly my teacher doesn't want to teach it to me. Can someone help me with these and explain. I would appreciate any help anyone can give me.

By Tanman (Tanman) on Sunday, September 28, 2003 - 11:29 pm: Edit

a) the numerator is a difference of 2 cubes - use the form (x^3-y^3) = (x-y)(x^2+xy+y^2):
(x-5)(x^2+5x+25). Then cancel the (x-5) to get (x^2+5x+25)

b) Plug 5 into the result from part a:
(5)^2+5(5)+25 = 75

By Bart_Simpson22 (Bart_Simpson22) on Sunday, September 28, 2003 - 11:35 pm: Edit

Thanks for the quick reply. However, I dont think I'm following you on problem a.). Could you eplain? I could be on the verge of understanding this though, thank you.

By Bart_Simpson22 (Bart_Simpson22) on Sunday, September 28, 2003 - 11:39 pm: Edit

anyone for c or d?

By Tanman (Tanman) on Sunday, September 28, 2003 - 11:46 pm: Edit

For (x^3-125), I quickly tell that both x^3 and 125 are perfect cubes. Therefore, I can use the special rules for factoring for difference of cubes which is any (x^3-y^3) can be factored into (x-y)(x^2+xy+y^2). In your case, "y" is equal to 5, so I plug this in to get (x-5)(x^2+x(5)+(5)^2). Then the (x-5)'s cancel out - this is the discontunity - and you are left with (x^2+5x+25). Hope that helps. If you're still having trouble, trying Googling "difference of two cubes"

By Bart_Simpson22 (Bart_Simpson22) on Sunday, September 28, 2003 - 11:57 pm: Edit

THANK YOU!!!! That really makes sense. What do you know about c and d?

By Bart_Simpson22 (Bart_Simpson22) on Monday, September 29, 2003 - 12:38 am: Edit

Anyone else?

By Jason817 (Jason817) on Monday, September 29, 2003 - 01:12 am: Edit

goddamit I forgot the intermediate value theorm

By Bart_Simpson22 (Bart_Simpson22) on Monday, September 29, 2003 - 07:05 am: Edit

bump

By Bart_Simpson22 (Bart_Simpson22) on Monday, September 29, 2003 - 01:20 pm: Edit

BUMP!

By Jshifton (Jshifton) on Wednesday, October 01, 2003 - 12:49 am: Edit

I had these exact same problems on my test from yesterday. Where did you get them from?

Here's how you do the intermediate value theorm problems...

First establish that f(x) is continous for the closed interval (a,b). Then calculate what f(a) and f(b) equal. If your number is between f(a) and f(b) there is a number x that equals it

The intermediate value theorm tells us this through common sense.

Now for number d.
Keep in mind that all polynomials are continous.
So you've established that. F(2)=16 F(3)=81. So any number between 16 and 81 has a value Y that equals a X between 2 and 3. coincedentally the 4th foot of 41=2.6723.

For number C. We know that all cosine functions are continous. So that has been established. Then...
F(0)= 1 for cosx, and F(90 degrees)=0. To show that X=cosX has at least one solution, we must simply match up x=x between the boundary's 0 and 1. Therefore there must be at least one solution between 0 and 1 for cosx=x. If you use your calculator it's roughly .74.

If you have questions, i'll gladly answer

By Jshifton (Jshifton) on Wednesday, October 01, 2003 - 10:00 pm: Edit

bump

By Bart_Simpson22 (Bart_Simpson22) on Saturday, October 04, 2003 - 04:17 pm: Edit

these were part of a practice test... thanks for the help.