# Thread: When the second derivative test fails

1. ## When the second derivative test fails

Hi,
My teacher has asked us to find the local extrema using the second derivative test. However I find one of my critical values fails the second derivative test. I have one example of this in my notes. In the example, the value that fails the second derivative test ends up not being a local minimum or maximum and we use the results from the second derivative test. However, in the problem I am working on now it turns out based on the first derivative test that the value that fails the second test is in fact the only local extrema, a local maximum. Because the value that failed the second test ended up being an extrema based on the first test do I ignore the results of the second test? Because looking at the graph the results of the second test are incorrect but the results of the first are.

the original function is:

f(x)= (9-x2)1/2

the work I've done:

first derivative: f'(x)= -x/(9-x2)1/2

second derivative: f''(x)=-x2/(9-x2)3/2

critical values: x= 1, -(3)1/2 , (3)1/2

results of the second derivative test:

f''(-(3)1/2)= -.20412 local maximum
f''(0)= 0 Local minimum
f''(31/2)= -.20412 local maximum

due to this result I continued with the first derivative test

Intervals: (-infinity, -(3)1/2) - increasing, (-(3)1/2, 0) - increasing, (0, 31/2) - decreasing, (31/2, infinity) -decreasing

results of the first derivative test:
x=0 yields a local maximum

f(0)=3

the graph of the original function is a half circle with points at (-3,0), (0,3), (3,0) hence the results of the first derivative test appear to be the accurate ones. So do I ignore the results from the second derivative test? and if so, is this the case any time the second derivative test fails and the value that fails is an extrema?

Thanks for the help.
I hope I provided enough clear info.

2. ## Re: When the second derivative test fails

Originally Posted by deviantxane
Hi,
My teacher has asked us to find the local extrema using the second derivative test. However I find one of my critical values fails the second derivative test. I have one example of this in my notes. In the example, the value that fails the second derivative test ends up not being a local minimum or maximum and we use the results from the second derivative test. However, in the problem I am working on now it turns out based on the first derivative test that the value that fails the second test is in fact the only local extrema, a local maximum. Because the value that failed the second test ended up being an extrema based on the first test do I ignore the results of the second test? Because looking at the graph the results of the second test are incorrect but the results of the first are.

the original function is:

f(x)= (9-x2)1/2

the work I've done:

first derivative: f'(x)= -x/(9-x2)1/2

second derivative: f''(x)=-x2/(9-x2)3/2

critical values: x= 1, -(3)1/2 , (3)1/2

results of the second derivative test:

f''(-(3)1/2)= -.20412 local maximum
f''(0)= 0 Local minimum
f''(31/2)= -.20412 local maximum

due to this result I continued with the first derivative test

Intervals: (-infinity, -(3)1/2) - increasing, (-(3)1/2, 0) - increasing, (0, 31/2) - decreasing, (31/2, infinity) -decreasing

results of the first derivative test:
x=0 yields a local maximum

f(0)=3

the graph of the original function is a half circle with points at (-3,0), (0,3), (3,0) hence the results of the first derivative test appear to be the accurate ones. So do I ignore the results from the second derivative test? and if so, is this the case any time the second derivative test fails and the value that fails is an extrema?

Thanks for the help.
I hope I provided enough clear info.
note the domain of the original function.

... also note that the critical values at $\displaystyle x = \pm 3$ are endpoints of a semicircle of radius 3.

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# second derivative test fails

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