The Second Derivative Test

The Second Derivative Test#

Using the Second Derivative to Classify Critical Points#

The Second Derivative Test

Suppose \(c\) is a critical point of \(f\) where \(f'(c) = 0\) and \(f''(x)\) is continuous near \(x=c\).

  1. If \(f''(c) > 0\), then \(f(x)\) has a relative minimum at \(x=c\).

  2. If \(f''(c) < 0\), then \(f(x)\) has a relative maximum at \(x=c\).

  3. If \(f''(c) = 0\), then the test is inconclusive and the First Derivative Test should be used instead.

Example 1#

Classifying critical points using the second derivative test

Find the relative extrema of \(f(x) = 12 x^5 - 45 x^4 - 200 x^3 +12\).

Step 1:   Find the critical points of   \(f\).
\[\begin{align*} f'(x) &= 60 x^4 - 180 x^3 - 600 x^2 \\ &= 60x^2(x^2-3x-10) && \text{pull out common factor of $60x^2$}\\ &= 60x^2(x-5)(x+2) && \text{factor}\\ \end{align*}\]

which equals zero when \(x=-2\), \(x=0\), and \(x=5\). Also note that \(f'(x)\) exists for all \(x\) in the domain of \(f\). Since the domain of \(f\) is \((-\infty,\infty)\), all of these values of \(x\) are critical points.

Step 2:   Compute   \(f''(x)\).
\[\begin{align*} f''(x) &= \frac{d}{dx}(60 x^4 - 180 x^3 - 600 x^2)\\ &= 240 x^3 - 540 x^2 - 1200 x \\ &= 60 x (4 x^2 - 9 x - 20) \\ \end{align*}\]
Step 3:   Evaluate   \(f''(x)\)   at each critical point.

Evaluate \(f''(x)\) at each critical point where \(f'(x)=0\) and use the Second Derivative Test to classify each critical point, if possible.

  • \(x=-2\): Since \(f'(-2) = 0\) and \(f''(-2) = 60(-2)(16+18-20) < 0\), \(f\) has a relative maximum at \(x=-2\).

  • \(x=0\): Since \(f'(0) = 0\) and \(f''(0) = 60(0)(-20) = 0\), the Second Derivative test is inconclusive. Instead, using the First Derivative test, we see there is no change in sign of \(f'(x)\) at \(x=0\), and therefore, \(f\) does not have a relative extreme at \(x=0\).

../_images/pic_curvesketching_secondderivativetest.png

Fig. 6 Interval analysis of \(f''(x) = 60 x (4 x^2 - 9 x - 20)\).#

Long Text Description

A number line with positive and negative signs assigned to intervals, with negative between negative two and zero, and negative between zero and five.

  • \(x=5\): Since \(f'(5) = 0\) and \(f''(5) = 60(5)(100-45-20) > 0\), \(f\) has a relative minimum at \(x=5\).

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