calc help please =/
 

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

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.

doh. thanks on that one =X

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.

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.

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.