Calculus

[UserLoser.]

I don't know how to do this or where to start, can someone show me?

Find the location of all horizontal tangent(s) to y=x+sin(x+1) in a domain of [0,2pi]. Give the (x,y) coordinates of your answer(s) and the corresponding tangent line equations.

[K]

Whever in doubt in calculus, differentiate.  The derivative of a function at a point is the slope at that point; by using the slope you can very easily write an equation for a tangent line.

d/dx(x + sin(x+1)) = 1 + cos(x + 1).

So that's the equation for the slope at any point x in the interval.  Then they tell you to look for horizontal tangents; ie, slope = 0;

Solve:
1 + cos(x + 1) = 0 on the close interval 0, 2pi.
1 = cos(x + 1)
cos(x) == 1 when x == pi; therefore, x = pi - 1.

This is your x coordinate. plug it back into the original equation to find the y coordinate.  then write an equation for a horizontal line that goes through that point.

[UserLoser.]

Hey, thanks.  You just got my brother who was too shy to post on these forums an A on his take home test   Of course, I don't know where to start though, so what I said was perfectally legitimate

[Lenny]

1 = cos(x + 1)
cos(x) == 1 when x == pi; therefore, x = pi - 1.

-1 = cos(x+1)*
cos(x) == -1 when x == pi

[Maddox]

Whever in doubt in calculus, differentiate.  The derivative of a function at a point is the slope at that point; by using the slope you can very easily write an equation for a tangent line.

d/dx(x + sin(x+1)) = 1 + cos(x + 1).

So that's the equation for the slope at any point x in the interval.  Then they tell you to look for horizontal tangents; ie, slope = 0;

Solve:
1 + cos(x + 1) = 0 on the close interval 0, 2pi.
1 = cos(x + 1)
cos(x) == 1 when x == pi; therefore, x = pi - 1.

This is your x coordinate. plug it back into the original equation to find the y coordinate.  then write an equation for a horizontal line that goes through that point.

cos(x) == 1 when x == 0 and when x == 2pi

[R.a.B.B.i.T]

cos(x) == 1 , x = npi && n == integer

[Yoni]

cos(x) == 1 , x = npi && n == integer

Incorrect;
x = 2n*pi && n is an integer.

Eww @ bringing up month old topic.

[Joe[x86]]

Hi Yoni. Shoot anybody lately?

Back on topic.. I hope Calculus isn't mandatory.

[Yoni]

no.

Calculus is mandatory. The universe wouldn't exist without it.

[R.a.B.B.i.T]

cos(x) == 1 , x = npi && n == integer

Incorrect;
x = 2n*pi && n is an integer.

Eww @ bringing up month old topic.

eww@dammit.  It was <lies>late</lies>, forgive me.

[Joe[x86]]

no.

Calculus is mandatory. The universe wouldn't exist without it.

Wow, I was almost certain that calculus wasn't mandatory. I guess it'll get easier as I learn it though. How hard can it really be if you have someone teaching you. Also, was that no meaning you haven't shot anybody lately? How's your military life going, anyhow?

[Adron]

Calculus is mandatory. The universe wouldn't exist without it.

Of course it would. The universe could get along just fine with just empirical data. No need for exactness.

[Yoni]

No. Calculus does not describe the universe; rather, the universe is an implementation of calculus.

Eww @ abusing admin powers to reply to locked topic (possibly by accident).

[Adron]

No. Calculus does not describe the universe; rather, the universe is an implementation of calculus.

Eww @ abusing admin powers to reply to locked topic (possibly by accident).

Hmm. Was it locked? I mostly notice the warning, but I have no recollection of it appearing for that reply.

And also, calculus is purely theoretic. You could never actually implement calculus, because calculus doesn't have an implementation. It's like philosophy or logic. You may implement something that somewhat follows rules from calculus, but any implementation is never really calculus.