Question

Sketch the graph of the function that has the following properties. f is continuous on (-infinity, infinity).
points: (-1,2), (0, 0), (-1,0)
f'(x)>0 at (-infinity, -1)
f'(-1)=0
f'(x)<0 at (-1, 1)
f'(1)=0
f'(x)>0 on (1, infinity)

f"(x)<0 on (-infinity, 0)
f"(0)=0
f"(x)>0 on (0, infinity)

I'm having trouble graphing with the given points because the critical points are different than the given points.
I do know:
points: (-1,2), (0, 0), (-1,0)
f increases on (-infinity, -1) U (1, infinity)
f decreases on (-1, 1)
critical points (-1,0) and (1,0)
f concave up on (0, infinity)
f concave down on (-infinity, 0)
inflection points on (0, 0)

Answer 1

both (-1,2) and (-1,0) are on the graph?

Is x a function of y?

Answer 2

Yes, I was confused why x=-1 and x=1 had two different points. Maybe it's a typo?

Answer 3

Wait the Points are (-1,0) (0,0) and (1,0). Im confused since the critical points and these points are both at x=-1, x=1

Answer 4

Sure looks like a typo to me.

Given that f'=0 at x = -1 and x=1,
f'(x) = k(x-1)(x+1) = x^2-1

That would make this just a plain old cubic curve, concave down for x<0 and concave up for x>0.

Then things fall into place and I think a good graph would be

y = x^3-3x

putting (1,-2) on the curve. Everything else fits.

http://www.wolframalpha.com/input/?i=x%5E3-3x

Answer 5

Okay thanks! I have drawn my graph like that!

Answer 6

Hmmm. So the curve is tangent to the x-axis at (-1,0) and (1,0) and has a point of inflection at (0,0)?

I don't see how that can be. The critical points work, but they cannot also be points on the curve.

It calls for too many changes in direction. I suspect the problem has been garbled.

Answer 7

Yeah That's where I was confused.Probably. Thank you!