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The Second Derivative TestPreviously, we determined whether a critical number $c$ of a
function $f$ indicated a local extreme value of $f$ by looking at
the sign of $f'$ on either side of $c$. If $f'$ changed sign
at $c$, there was a local extreme value there, since $f'$ changes
sign exactly where $f$ changes direction. This is the first derivative test, where we use the
first derivative to determine where we have local extrema. On this slide we use another method that might allow us to
determine whether there is a local extreme value at $x=c$.
It uses the second derivative as well as the first, so we call it
the second derivative test. First, we formalize the concavity concepts from our previous
work:
Another perspective: If $f''(c) \gt 0$, then $f'(x)$ is
increasing near $x=c$. Since $f'(c)=0$, this means that $f'(x)$
used to be negative and is about to be positive. So the curve
bottoms out at $x=c$ and then heads back up. The critical number
$x=c$ is the bottom of the concaveup bowl. Likewise, if $f''(c)
\lt 0$ and $f'(c)=0$, then $f'(x)$ is decreasing; it used to be
positive and is about to be negative. The point $x=c$ is at the
top of an upsidedown bowl. Notice: As is indicated in the third option of the test, if a critical number $c$ is also a subcritical
number, then the second derivative test cannot help determine
whether or not there is a max or min at $c$.
DO: Try this before reading the solution, using the process above.
To use the second derivative test, we check the concavity of $f$ at the critical numbers. We see that at $x=0$, $x<1$ so $f$ is concave down there. Thus we have a local maximum at $x=0$. At $x=2$, since $x>1$ $f$ is concave up there, so we have a local minimum at $x=2$. These results agree with what we got from the first derivative test.
