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5.7 Using the Second Derivative Test to Determine Extrema

Syllabus
2020
Topic
5.7
Level
—

Learning objectives

—FUN-4.A—Justify conclusions about the behavior of a function based on the behavior of its derivatives—Topic 5.7• FUN-4.A Justify conclusions about the behavior of a function based on the behavior of its derivatives.• FUN-4.A.7 The second derivative of a function may determine whether a critical point is the location of a relative (local) maximum or minimum.• FUN-4.A.8 When a continuous function has only one critical point on an interval on its domain and the critical point corresponds to a relative (local) extremum of the function on the interval, then that critical point also corresponds to the absolute (global) extremum of the function on the interval.• Enduring understanding FUN-4: A function’s derivative can be used to understand some behaviors of the function.

Classify a Critical Point with $f''$

The Second Derivative Test uses concavity to classify a stationary critical point. First verify that f′(c)=0f'(c)=0 and that f′′(c)f''(c) exists. Then the sign of f′′(c)f''(c) tells whether the graph bends upward or downward at x=cx=c.

Value of f′′(c)f''(c) Concavity near cc Conclusion
f′′(c)>0f''(c)>0 concave up local minimum at x=cx=c
f′′(c)<0f''(c)<0 concave down local maximum at x=cx=c
f′′(c)=0f''(c)=0 not determined test is inconclusive

Solve f′(x)=0f'(x)=0 for the stationary critical points, calculate f′′(c)f''(c) at each one, and state the sign-based conclusion. If f′′(c)=0f''(c)=0 or does not exist, use another method such as the First Derivative Test; do not force a classification.

For f(x)=x2+4x+1f(x)=x^2+4x+1, f′(x)=2x+4f'(x)=2x+4, so the only critical point is c=−2c=-2. Since f′′(x)=2>0f''(x)=2>0, ff has a local minimum at x=−2x=-2. The polynomial is continuous on (−∞,∞)(-\infty,\infty) and has only this one critical point, so the local minimum is also the absolute minimum; its value is f(−2)=−3f(-2)=-3.

The global conclusion needs all of its own conditions: continuity on the interval, exactly one critical point there, and proof that this point is a local extremum. Also, f′′(c)=0f''(c)=0 does not mean “no extremum”; it means this test gives no answer.