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 given. 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 http://mathforum.org/dr.math/ 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! |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/