calc help please =/

[fuzzybottom]

two questions im stuck on, don't have access to the teacher atm so i can't ask until tomorrow.

graph of f`, derivative of f, has a domain of 0>x>2 and since i cant draw it
here: its a straight line from (0,0) to (1,1) then a straight line from (1,1)
to (2,0) and it exists at (1,1).
if it asks for an expression for f`(x), the graph they give, in terms of x,
do you guys think they want the set of equations with domain restrictions that would cause that, or what? i end up having to write the expression for f(x) after that given f(1) = 0, i can do that part, but i dont know what to
do about the first part.
if its the set of equations i've got:
f`(x) = x, 0 < x < 1
f`(x) = 1, x=1
f`(x) = 2-x, 1 < x < 2

and does anyone know about vertical tangent lines for x² + xy + y² = 27
im still not sure about questions like this, whether there is a vertical tangent line or what, since when i take the derivative i get
dy/dx = -(2x + y)/(x + 2y)
and for there to be a veritical tangent line that would have to be undefined at some point, so x must be -2y, and x=-2y has infinite solutions =/

thanks for any help you can give

[a-j]

x=-2y does have infinitely many solutions, but only a few solutions given your initial conditions, namely: x^2+xy+y^2=27. Simple substitution should give you your answers.

[fuzzybottom]

doh. thanks on that one =X

[C4 Diesel]

Originally Posted by fuzzybottom

f`(x) = x, 0 < x < 1
f`(x) = 1, x=1
f`(x) = 2-x, 1 < x < 2

Take out that middle equation (f`(x) = 1, x=1) and replace the "x < 1" in the first equation with x (is grater than or equal to) one, and do the same for the first half of the last equation (so one is less than or equal to x ...). 1,1 is not defined because the the y-value happens to be 1 at x=1, but rather because the other equations are equal to one at that point.

[fuzzybottom]

hmm k, i'll try that.
btw do you think thats the best way to go about that problem? i tried to use:
-[(x-1)²]^(1/2) +1
and integrate that but it got messy, and i'm not sure if thats a good way to go about it because of the exponents and how they might distribute. i guess the first way would be easier.

[a-j]

Originally Posted by fuzzybottom

i tried to use:
-[(x-1)²]^(1/2) +1
and integrate that but it got messy, and i'm not sure if thats a good way to go about it because of the exponents and how they might distribute.

I guess that would work, but it is ugly. I think using abolute values would be much cleaner:

f'(x) = -|x-1|+1

Then if you need to integrate it it would break down into pieces:

x: 0&lt;x&lt;=1
-x+2: 1&lt;x&lt;2

Which is what you have already. It really isn't that uncommon for integrals to be evaluated piecewise, and it makes your job a lot easier than messing around trying to get one nice function. After all the function isn't smooth, so I wouldn't try to integrate it as if it were.