Expert: Paul Klarreich - 11/9/2010

Question
Sketch a graph that satisfies all the given conditions:

f'(x) > 0 if |x|<2
f'(x) < 0 if |x|>2
f'(2) = 0
the limit as x->infinity of f(x) = 1
f(-x) = -f(x)
f"(x) < 0 if 0 < x < 3
f"(x) > 0 if x > 3

I have concluded that the graph is odd
and that at x = 2 there is neither a maximum
nor a minimum. I don't know how to interpret
the rest of the conditions.

Thank you in advance.

Answer
Questioner:   Robert
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  Sketching a curve.
Question:  Sketch a graph that satisfies all the given conditions:

f'(x) > 0 if |x|<2
f'(x) < 0 if |x|>2
f'(2) = 0
the limit as x->infinity of f(x) = 1
f(-x) = -f(x)
f"(x) < 0 if 0 < x < 3
f"(x) > 0 if x > 3

I have concluded that the graph is odd and that at x = 2 there is neither a maximum
nor a minimum. I don't know how to interpret
the rest of the conditions.

Thank you in advance.
-------------------------------------
Well, let's draw some conclusions.  Then you'll draw some pictures.

First we focus on positive x-values. (You'll see why later.)

f'(x) > 0 if |x|<2

-- the graph rises on (0,2)

f'(x) < 0 if |x|>2

-- the graph falls on (2,inf)

f'(2) = 0

-- horizontal tangent at  x = 2.

WE HAVE A MAXIMUM AT x = 2.

the limit as x->infinity of f(x) = 1

-- horizontal asymptote is  y = 1, and the graph is gently sliding down to it.  (sorry for the colorful phrasing)

f(-x) = -f(x)

-- f is an odd function, so its graph has origin symmetry, which means the left side looks like the right side, but inverted.

f"(x) < 0 if 0 < x < 3

-- concave downward on (0,3)

f"(x) > 0 if x > 3

-- concave upward on (3,inf)


I have concluded that the graph is odd

-- NO NO NO! The FUNCTION is odd, and the graph is symmetric.


and that at x = 2 there is neither a maximum
nor a minimum.

-- Not so; see above.

This should do it.  If you need more help, let me know.