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Can f'(-1) Equal Zero and f''(-1) Not Equal Zero?

Date: 03/23/2004 at 17:05:30
From: Elizabeth
Subject: f'(-1)=0 but f''(x)= a negative # is this possible?

In calculus, we are exploring sketching graphs and determining 
critical points using number lines that tell us values at certain 
points on the graphs of f'(x) and f''(x).  The original f(x) is not 

In one problem, the chart defines f'(-1) to be 0 and then f''(-1) is
negative.  I didn't think it was possible to have a derivative of zero
and then have a double derivative that is not zero at that same x
value.  Is this possible, and if so could you give me an example of
how this would work? 

I thought that when you take a derivative and it is equal to zero at
some point then the second and third and so on derivatives would
always be equal to zero at that point.

Date: 03/23/2004 at 17:25:40
From: Doctor Achilles
Subject: Re: f'(-1)=0 but f''(x)= a negative # is this possible?

Hi Elizabeth,

Thanks for writing to Dr. Math.

That's an excellent and thoughtful question.

You are right, that for the derivative to be zero, the slope (and 
therefore the tangent line) must be zero.

But what exactly does a double derivative measure?

Well, a single derivative measures the slope, or rate of change in 
the y-value of the graph.  So a double-derivative must measure the 
rate of change of the slope.  When the double-derivative is positive, 
that means that the slope is increasing at that point, and when it is 
negative it means the slope is decreasing.

Just like you can have a function whose y-value is 0, but whose slope 
is not, you can have a function whose slope is 0 but whose slope is 
constantly changing and therefore never has a double-derivative of 0.

Take this example:

  f(x) = x^2
  f'(x) = 2x
  f''(x) = 2

You'll notice here that the derivative is 0 at x = 0, but the double-
derivative is always 2, which means that the slope is always increasing.

You may recognize that function as a basic upwards opening parabola
with vertex at (0,0).  Let's think about what each statement is
telling us at x = 0.  Since f(0) = 0, the graph of the function hits
the origin.  Since f'(0) = 0 the slope of the graph as it passes
through the origin is 0.  You can visualize that the tangent to the
parabola at the origin would essentially be the x-axis, with slope of 0.  

Now, since f''(0) = 2, we know that the slope of the parabola is
increasing.  If you visualize the tangent lines to the parabola, they
start out with a very large negative slope on the left side.  The
slopes becomes less negative as we move along the x-axis, and finally
the tangent flattens out right at the origin.  Then, as we continue to
the right, the tangent lines start to become steeper again, but now in
a positive slope.  The net result of all this is that the slopes of
the tangents, or of the parabola itself, consistently increase, moving
from a very large negative up to 0 and then onwards to a very large
positive.  The fact that f''(x) = 2 tells us that the slope of the
curve is constantly increasing.

For an example that could give the exact behavior of the problem you 
were assigned, calculate the derivatives and double-derivatives of 
these, and evaluate them at x = -1.

  g(x) = x^3/3 - x

  h(x) = - x^2/2 - x

I hope this is helpful.  If you'd like to discuss this more, please 
write back.

- Doctor Achilles, The Math Forum 

Date: 03/23/2004 at 21:08:33
From: Elizabeth
Subject: Thank you (f'(-1)=0 but f''(x)= a negative # is this possible?)

Thank you so much for your quick reply.  I realize now that I was
thinking about the problem wrong and I really appreciate your help
because now I get it.  Thanks!
Associated Topics:
High School Calculus
High School Functions

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