Second derivative test

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Here we look at the second derivative test.

Using the second derivative to locate extrema

Recall the first derivative test:

If $f'(x)>0$ to the left of $a$ and $f'(x)<0$ to the right of $a$ , then $f(a)$ is a local maximum.
If $f'(x)<0$ to the left of $a$ and $f'(x)>0$ to the right of $a$ , then $f(a)$ is a local minimum.

If $f'$ changes from positive to negative it is decreasing. In this case, $f''$ might be negative, and if in fact $f''$ is negative then $f'$ is definitely decreasing, so there is a local maximum at the point in question. On the other hand, if $f'$ changes from negative to positive it is increasing. Again, this means that $f''$ might be positive, and if in fact $f''$ is positive then $f'$ is definitely increasing, so there is a local minimum at the point in question. We summarize this as the second derivative test.

Second Derivative Test Suppose that $f''(x)$ is continuous on an open interval and that $f'(a)=0$ for some value of $a$ in that interval.

If $f''(a) <0$ , then $f$ has a local maximum at $a$ .
If $f''(a) >0$ , then $f$ has a local minimum at $a$ .
If $f''(a) =0$ , then the test is inconclusive. In this case, $f$ may or may not have a local extremum at $x=a$ .

The second derivative test is often the easiest way to identify local maximum and minimum points. Sometimes the test fails and sometimes the second derivative is quite difficult to evaluate. In such cases we must fall back on one of the previous tests.

Once again, consider the function $f(x) = \frac {x^4}{4}+\frac {x^3}{3}-x^2$ Use the second derivative test, to locate the local extrema of $f$ .

Start by computing $f'(x) = \answer [given]{x^3 + x^2 -2x} \qquad \text {and}\qquad f''(x) = \answer [given]{3x^2 + 2x-2}.$ Using the same technique as we used before, we find that $f'(-2) = \answer [given]{0},\qquad f'(0) = \answer [given]{0}, \qquad f'(1) = \answer [given]{0}.$ Now we’ll attempt to use the second derivative test, $f''(-2) = \answer [given]{6}, \qquad f''(0) =\answer [given]{ -2}, \qquad f''(1) = \answer [given]{3}.$ Hence we see that $f$ has a local minimum at $x=-2$ , a local maximum at $x=0$ , and a local minimum at $x=1$ , see below for a plot of $f(x) =x^4/4 + x^3/3 -x^2$ and $f''(x) = 3x^2 + 2x -2$ :

If $f''(a)=0$ , what does the second derivative test tell us?

The function has a local extrema at $x=a$ . The function does not have a local extrema at $x=a$ . It gives no information on whether $x=a$ is a local extremum.