NationStates Jolt Archive

I'm currently doing a project for Algebra II and I just realized the teacher wouldn't be there tomorrow (Monday; Project due Tuesday) so I wouldn't be able to ask her about this problem.

Suppose m and k are numbers and let g(x) be defined piecewise as follows:

g(x) = { 1/2x + k if x<3
mx+2 if x is greater than or equal to 3.



The first question:

Suppose m=3/2. What must the value of k be in order for the graph of y=g(x) to be connected?

There's a second question but if someone could point me in the right direction with the first I can take care of the second.

I'm pretty sure I could help with this, but your syntax isn't making much sense to me.

do you mean

if x<3 then g(x) = 1/2x + k
if x> then g(x) = mx+2

if m = 3/2 then what must the value of k be if y = g(x) is connected (that's strangely phrased, but I get it)

Did the '{' mean anything in g(x) = { 1/2x + k ?

Basically I think the question is find k when m= 3/2 and x= a constant (you have to figure what that constant is)

I'm currently doing a project for Algebra II and I just realized the teacher wouldn't be there tomorrow (Monday; Project due Tuesday) so I wouldn't be able to ask her about this problem.

Suppose m and k are numbers and let g(x) be defined piecewise as follows:

g(x) = { 1/2x + k if x<3
mx+2 if x is greater than or equal to 3.



The first question:

Suppose m=3/2. What must the value of k be in order for the graph of y=g(x) to be connected?

There's a second question but if someone could point me in the right direction with the first I can take care of the second.
Just equate the two functions of g(x) and solve for k. These are just linear equations. Figure out where they intersect.

I could help you if I really put forth a lot of effort, but we did picewise functions in the middle of last semester and since I already took the final, I've forgotten it already.

Sorry if the question wasn't clear.

if x<3 then g(x) = 1/2x + k
if x> then g(x) = mx+2

if m = 3/2 then what must the value of k be if y = g(x) is connected

That's it; the { doesn't mean anything

After playing some RTW, I tried out the equation:

1/2x+k = 3/2x + 2

Reduced it finally to:

k = x + 2

I plugged in 3 for x and on the graph it looks good, but I'm not sure if that's the right way of doing it.

you see, I would have plugged the 3 in beforehand, but I got 5 as well. Right way, way wrong? You took the right infomation and got the right answer. As far as I'm concerned it's the right way.

I just did peicewise functions

you see, I would have plugged the 3 in beforehand, but I got 5 as well. Right way, way wrong? You took the right infomation and got the right answer. As far as I'm concerned it's the right way.

Thanks.

(3/2) * 3 + 2 = lim(x-->3) (1/2) * x + k

They connect where x = 3. So plug in 3 for x on the side where x is included, have x approach 3 on the side where it's excluded, and solve for k

edit: You're lucky to have gotten the right answer by plugging in 3 for x on the side where it's excluded. That doesn't always work

I'm currently doing a project for Algebra II and I just realized the teacher wouldn't be there tomorrow (Monday; Project due Tuesday) so I wouldn't be able to ask her about this problem.

Suppose m and k are numbers and let g(x) be defined piecewise as follows:

g(x) = { 1/2x + k if x<3
mx+2 if x is greater than or equal to 3.

The first question:

Suppose m=3/2. What must the value of k be in order for the graph of y=g(x) to be connected?

There's a second question but if someone could point me in the right direction with the first I can take care of the second.

k=5

Plug 3 in for x and 3/2 in for m in the second half of the equation. Solve to find the value of g. Set the first half of the equation to this value and plug in 3 for x. Solve for k.

Ok.

g(x)= 1/2x + k for x < 3
mx + 2 for x >= 3

If m = 3/2, then at x = 3, g(x) = (3/2)(3) + 2 = 13/2. Now just solve

(1/2)(3) + k = 13/2.

3/2 + k = 13/2.

k = 13/2 - 3/2 = 10/2 = 5.

k = 5.

edit: You're lucky to have gotten the right answer by plugging in 3 for x on the side where it's excluded. That doesn't always work

It's not luck, it's a linear function. The rule for finding the limit is to "plug in."

It's not luck, it's a linear function. The rule for finding the limit is to "plug in."and if the function happens to be sin[x] / x, with the boundary at x = 0?

and if the function happens to be sin[x] / x, with the boundary at x = 0?

Then it wouldn't be linear, would it?

EDIT: And actually, the "plug it in" rule works for polynomials in general.

Sure, you could pull out epsilon and prove the limit directly... but since no one ordinarily wants to do that, we have the simple rule: plug it in.

In algebra, we are nice to the little kids and don't give them anything else to worry about. So it's not "luck." It's the fact that the people who write the math book know it's going to work out... and the method for getting there is perfectly valid, even in calculus.

Sorry if the question wasn't clear.


That's it; the { doesn't mean anything

After playing some RTW, I tried out the equation:

1/2x+k = 3/2x + 2

Reduced it finally to:

k = x + 2

I plugged in 3 for x and on the graph it looks good, but I'm not sure if that's the right way of doing it.

for the function to be continuous, lim(g, x->3, x<3) = lim(g, x->3, x>3) = lim(g, x->3). as taking the limit of those functions is just replacing x by 3, you end up with k = 5. No magic done.

and if the function happens to be sin[x] / x, with the boundary at x = 0?

l'Hopital:

lim(sin(x)/x,x=0) = 0/0
=> l'Hopital lim((d(sin(x))/dx)/(dx/dx),x=0) = lim(cos(x)/1,x=0) = 1

Quit hammering the poor guy with Calculus, he'll overload. :D

Btw, anybody know any tricks for matrix multiplication? I'm taking LinAlg and some of those problems can get pretty long.

d(whole bunch of crap)/dx=your head a splode!

l'Hopital:

lim(sin(x)/x,x=0) = 0/0
=> l'Hopital lim((d(sin(x))/dx)/(dx/dx),x=0) = lim(cos(x)/1,x=0) = 1
Unfortunately, in order to find the derivative of the sine, you need to know the value of that limit. You actually need to apply the squeeze theorem to 1 and cos x.

Quit hammering the poor guy with Calculus, he'll overload. :D

Btw, anybody know any tricks for matrix multiplication? I'm taking LinAlg and some of those problems can get pretty long.
I use a calculator.

I think L'Hopitals Rule only applies to Limits taken to infinity.

I was sure it applies for both, but i looked it up:

1 if f and g are derivable in ]a,b[ (a, b can be infinity), dg/dx is not 0 in ]a,b[ and lim(f,x->a) = 0 = lim(g,x->a) then:

1.1 if lim((df/dx)/(dg/dx),x->a) = L, then lim(f/g,x->a) = L
1.2 if lim((df/dx)/(dg/dx),x->a) = +-infinity, then lim(f/g,x->a) = +-infinity

2 if f and g are derivable in ]a,b[ (a, b can be infinity), dg/dx is not 0 in ]a,b[ and lim(f,x->a) = +-infinity = lim(g,x->a) then:

2.1 if lim((df/dx)/(dg/dx),x->a) = L, then lim(f/g,x->a) = L
2.2 if lim((df/dx)/(dg/dx),x->a) = +-infinity, then lim(f/g,x->a) = +-infinity

Unfortunately, in order to find the derivative of the sine, you need to know the value of that limit. You actually need to apply the squeeze theorem to 1 and sec x.

let's see.

the 'definition' of sin and cos we've seen and used:
e^z = sum(z^k,k=0..infinity)
sin(x) = i/2*(e^-ix-e^ix)
cos(x) = 1/2*(e^ix+e^-ix)

and with d(e^az)/dz = a*e^z

this allows me to take the derative of sin(x) i believe?

let's see.

the 'definition' of sin and cos we've seen and used:
e^z = sum(z^k,k=0..infinity)
sin(x) = i/2*(e^-ix-e^ix)
cos(x) = 1/2*(e^ix+e^-ix)

and with d(e^az)/dz = a*e^z

this allows me to take the derative of sin(x) i believe?
No, Euler's theorem also requires knowledge of the derivative of the sine. It comes from a uniqueness theorem about the differential equation f"(x) = -f(x), f(0) = 1, f'(0) = i.

I was sure it applies for both, but i looked it up:

1 if f and g are derivable in ]a,b[ (a, b can be infinity), dg/dx is not 0 in ]a,b[ and lim(f,x->a) = 0 = lim(g,x->a) then:

1.1 if lim((df/dx)/(dg/dx),x->a) = L, then lim(f/g,x->a) = L
1.2 if lim((df/dx)/(dg/dx),x->a) = +-infinity, then lim(f/g,x->a) = +-infinity

2 if f and g are derivable in ]a,b[ (a, b can be infinity), dg/dx is not 0 in ]a,b[ and lim(f,x->a) = +-infinity = lim(g,x->a) then:

2.1 if lim((df/dx)/(dg/dx),x->a) = L, then lim(f/g,x->a) = L
2.2 if lim((df/dx)/(dg/dx),x->a) = +-infinity, then lim(f/g,x->a) = +-infinity

I changed my post, I just had to think about it a while. Do you just have mad memory skill, or are you currently studying this?

I use a calculator.

Not allowed at UAH.

Not allowed at UAH.
Practice your arithmetic?

No, Euler's theorem also requires knowledge of the derivative of the sine. It comes from a uniqueness theorem about the differential equation f"(x) = -f(x), f(0) = 1, f'(0) = i.

Then i guess you are right. I've only seen what is 'usefull' for application in science, so i haven't seen a lot of theoremas that are used to prove the theorema's we use. But with l'Hopital proven, it's still a practical way of calculating the limit (allthough less practical than just using my computer), even with the circular logic.

I changed my post, I just had to think about it a while. Do you just have mad memory skill, or are you currently studying this?
Sigh. http://forums2.jolt.co.uk/showthread.php?p=8005047#post8005047

Then i guess you are right. I've only seen what is 'usefull' for application in science, so i haven't seen a lot of theoremas that are used to prove the theorema's we use. But with l'Hopital proven, it's still a practical way of calculating the limit (allthough less practical than just using my computer), even with the circular logic.
Well, you could just skip the l'Hopital part, then. Start deriving the derivative of the sine. When you get to cos x * (sin h)/h, then just state that you know the derivative of the sine. 8)

I understand the basics fairly well and I stil can't derive shit like that on the fly. Ny left brain has been suffering as of late, though. I'm not sure what it is, but I've been contiually getting dumber of the past few years.

Unfortunately, in order to find the derivative of the sine, you need to know the value of that limit.
No, we know the derivative of sin, and it has nothing to do with that limit. It's cos(x). Using l'Hopital's rule does not require that one proves every limit every time.

By the way, we prove the derivative of sin(x) as follows:

f(x) = sin(x)

f'(x) = lim(h->0) {[f(x+h)-f(x)]/h} by definition

= lim {[sin(x+h) - sin(x)]/h}

= lim {[sin(x)cos(h) + cos(x)sin(h) - sin x]/h}

= lim {sin(x)*[(cos(h)-1)/h] + cos(x)*[sin(h)/h]}

= lim sin(x)*lim[(cos(h)-1)/h] + lim cos(x)*lim[sin(h)/h]

= sin(x)*0 + cos(x)*1

= cos(x)

All using simply trigonometry.

No, we know the derivative of sin, and it has nothing to do with that limit. It's cos(x). Using l'Hopital's rule does not require that one proves every limit every time.

By the way, we prove the derivative of sin(x) as follows:
...
= lim sin(x)*lim[(cos(h)-1)/h] + lim cos(x)*lim[sin(h)/h]

= sin(x)*0 + cos(x)*1

Now explain how we didn't need to know the value of lim[sin(h)/h].

Yay, good ol' limit derivatives. Thank God for the power rule.

Now explain how we didn't need to know the value of lim[sin(h)/h].because the unit circle is vertical at the x-axis

because the unit circle is vertical at the x-axis
That's a start. I didn't say that the limit was impossible to prove. I said that l'Hopital's rule wouldn't do it.