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Sketching a Graph Given Information about Its Derivatives

Date: 07/30/2005 at 14:46:49
From: sahbina
Subject: Graphing a function when not given f(x), but characteristics

In class we are taking f'(x) and f''(x) and finding the intervals 
where f'(x) is increasing and decreasing and for f''(x) where the 
function is concave up and concave down, also any asymptotes.  But for 
this question we are given some info and we have to graph the 
function.  It's like going in reverse. For example,

  f(0)=4 and f(6)=0
  f'(x)<0 if x<2 and x>4
  f'(2) does not exist
  f'(x)>0 if 2<x<4
  f''(x)<0, x cannot=2

From the information I can see that {0,6) are the boundaries.  Also 
x=2,4 are points on the graph of f(x) because they make f'(x) and 
f''(x) equal 0.  Altogether f(0), f(2), f(4), and f(6) are all points 
of the graph of f(x).  My teacher told us f'(4)=0, but that is evident 
because 4 is an x value for f'(x) which makes it equal to 0.  I 
attempted to graph the function, but I feel like there is a point 
missing on the graph because f(4) does not show up on the graph, and 
it is indeed a point.

Also f(x) is decreasing at  (-infinity, 2) and ( 4, +infinity), and
f(x) is increasing at (2,4).  Also f''(x) is concave down at 
(-infinity,2) and (2, +inifinity). Also there is a vertical asymptote 
at x=2.  There has to be a point at f(4), but he didn't give it to us.

Date: 07/30/2005 at 23:05:27
From: Doctor Peterson
Subject: Re: Graphing a function when not given f(x), but characteristics

Hi, Sahbina.

I'm not sure what you mean by a few of your statements; your 
terminology is unclear sometimes.  Be careful: x=2 is not a point, 
it's just a value of x!  And f(0) is not a point, but a value of y. 
Let's look at the list of facts, and express them in English:

>f(0)=4 and f(6)=0        The points (0,4) and (6,0) are on the graph
>f'(x)<0 if x<2 and x>4   The graph is decreasing in the intervals
                          (-oo,2) and (4,oo)
>f'(2) does not exist     The graph has no tangent, or a vertical
                          tangent, at x=2
>f'(4)=0                  The graph is horizontal at x=4
>f'(x)>0 if 2<x<4         The graph is increasing in (2,4)
>f''(x)<0, x cannot=2     The graph is concave down everywhere except
                          where its derivative does not exist

There is no mention of boundaries; x can be anything.  Rather, 0 and 6
are special values of x for which y is specified.  Your comments on
increasing, decreasing, and concave down intervals are correct, but
you can't say there is a vertical asymptote at x=2, since then f(2)
would not exist.  (We would like to know f(2) and f(4), but since we
aren't told those, we will just have to guess.  It doesn't mean the
points don't exist!)

To sketch the graph, I would start by plotting the two specifically
known points, (0,4) and (6,0).  Then I would mark where the graph is
increasing and decreasing, perhaps by making a dotted vertical line at
x=2 and x=4 to mark the regions, and then drawing lightly a line
slanting downward in the decreasing intervals:

     \|       :       :
      o       :       :
      |\      :      /:\
      + \     :     / : \
      |  \    :    /  :  \
      +   \   :   /   :   \
      |    \  :  /    :    \
      +     \ : /     :     \
      |      \:/      :      \
      |                        \

That zigzag doesn't represent the actual graph, just one fact about
it, namely the direction of the slope.  (I happen to have made it pass
through the two given points, as well, which makes it fairly close to
a possible graph of f.)

Next, we can look at the details about the first derivative, namely
that it is zero at x=4 and doesn't exist at x=2.  We can draw a
horizontal tangent at x=4, and (since we aren't told that f(4) doesn't
exist) a vertical slope at x=2:

     \|       :       :
      o       :       :
      |\      :      /:\
      + \     :     / : \
      |  \    :    /  :  \
      +   \   :   /   :   \
      |    \  :  /    :    \
      +     \ : /     :     \
      |      \:/      :      \
      |                        \

Now we can check what is said about the second derivative: it's
negative everywhere it exists.  That means the graph is concave down.
We have to "bend" the lines so they fit that description:

  *   |       :      ***
      o       :   *   :  *
      | *     :       :   *
      +   *   : *     :    *
      |     * :       :  
      +      *:*      :     *
      |       :       :    
      +       *       :      *
      |       :       :

I made sure it was always concave down, and (though you can't see it!)
made it tangent to x=2 where it reaches it.  That is called a cusp.

None of the actual points except (0,4) and (6,0) are necessarily
correct, and the local minimum and maximum may be far from where they
really belong; but we've sketched a graph that fits all the 
information we were given, which is the goal.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum 
Associated Topics:
High School Calculus
High School Functions

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