AP Calculus AB/BC · Unit 5 · Topic 5.7
Using the Second Derivative Test to Determine Extrema
Classify a critical point using f″ when the test is conclusive.
1. Topic Focus
Use derivatives to prove existence, classify extrema, analyze monotonicity and concavity, sketch graphs, and solve optimization problems.
This topic: Classify a critical point using f″ when the test is conclusive.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
If f′(1)=0 and f″(1)<0, the graph is locally concave down and has a local maximum.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
1. What the test determines
The Second Derivative Test classifies a stationary critical number as a local maximum or minimum. It uses the graph's concavity near a horizontal tangent and is often faster than building a complete sign chart for \(f'\).
2. State the hypotheses
Suppose \(f'(c)=0\) and \(f''\) exists and is continuous on an open interval around \(c\). Then the sign of \(f''(c)\) can classify \(f(c)\). The condition \(f'(c)=0\) must be checked before using the test.
3. Why positive means minimum
If \(f''(c)>0\), then \(f'\) is increasing near \(c\). Since \(f'(c)=0\), nearby slopes move from negative to positive, so \(f\) decreases and then increases. Therefore \(f(c)\) is a local minimum.
4. Why negative means maximum
If \(f''(c)<0\), then \(f'\) is decreasing near \(c\). Nearby slopes move from positive to negative, so \(f\) increases and then decreases. Therefore \(f(c)\) is a local maximum.
5. The classification table
6. Use the correct order
- Find \(f'(x)\).
- Solve \(f'(x)=0\) for stationary critical numbers.
- Find \(f''(x)\).
- Evaluate \(f''\) at each stationary critical number.
- Classify when the result is positive or negative.
- If a point is requested, calculate \(f(c)\) and report \((c,f(c))\).
7. Zeros of the second derivative are not the candidates
Solving \(f''(x)=0\) locates possible concavity changes, not possible extrema for this test. Extrema candidates come first from \(f'(x)=0\) or from points where \(f'\) is undefined and \(f\) exists.
8. Nondifferentiable critical points need another test
A corner or cusp can be a local extremum, but the Second Derivative Test cannot classify it because \(f'(c)=0\) is not satisfied. Use the First Derivative Test or compare nearby function values instead.
9. Zero means inconclusive
If \(f'(c)=0\) and \(f''(c)=0\), no classification follows. At zero, \(x^4\) has a local minimum, \(-x^4\) has a local maximum, and \(x^3\) has neither. The same second-derivative value permits all three outcomes.
10. Undefined also means the test does not decide
If \(f''(c)\) does not exist, the test is unavailable, not evidence that no extremum exists. Return to the sign of \(f'\), the function's local behavior, or another valid argument.
11. Know when the First Derivative Test is better
The First Derivative Test works at stationary and nondifferentiable critical points and resolves cases where the Second Derivative Test is inconclusive. The second-derivative method is efficient when \(f''(c)\) is easy to evaluate and nonzero.
12. Local is not automatically absolute
The test gives a neighborhood conclusion. To find absolute extrema on a closed interval, also evaluate endpoints and every critical candidate using the Candidates Test. A local minimum can still be higher than an endpoint value.
13. Common errors
- Solving \(f''=0\) instead of \(f'=0\) for candidates.
- Using the test when \(f'(c)\ne0\).
- Calling \(f''(c)=0\) “neither” instead of inconclusive.
- Forgetting critical points where \(f'\) is undefined.
- Reporting a local result as an absolute result.
- Giving only \(c\) when the problem requests the point or value.
6. Detailed Worked Example and Error Check
Example 1: Several stationary critical numbers. Let \(f(x)=x^4-4x^2\).
The stationary critical numbers are \(0\) and \(\pm\sqrt2\). Since \(f''(0)=-8\), \((0,0)\) is a local maximum. Since \(f''(\pm\sqrt2)=16\), \((\pm\sqrt2,-4)\) are local minima.
Example 2: A cubic with one maximum and one minimum. For \(f(x)=x^3-3x^2\), \(f'(x)=3x(x-2)\) gives \(c=0,2\), and \(f''(x)=6x-6\). Thus \(f''(0)=-6\) gives a local maximum at \((0,0)\), while \(f''(2)=6\) gives a local minimum at \((2,-4)\).
Example 3: Odd-degree polynomial. Let \(f(x)=x^5-5x\). Since \(f'(x)=5(x^4-1)\), the real stationary critical numbers are \(-1\) and \(1\). With \(f''(x)=20x^3\), \(f''(-1)<0\) gives a local maximum at \((-1,4)\), and \(f''(1)>0\) gives a local minimum at \((1,-4)\).
Example 4: Three inconclusive outcomes. Each of \(x^4\), \(-x^4\), and \(x^3\) satisfies \(f'(0)=f''(0)=0\). The Second Derivative Test is inconclusive for all three. A First Derivative Test shows a minimum for \(x^4\), a maximum for \(-x^4\), and neither for \(x^3\).
Example 5: Exponential function. For \(f(x)=e^x-2x\), the equation \(f'(x)=e^x-2=0\) gives \(c=\ln2\). Since \(f''(\ln2)=e^{\ln2}=2>0\), the function has a local minimum at
Example 6: Trigonometric function. For \(f(x)=\sin x+\cos x\) on \((0,2\pi)\), \(f'(x)=\cos x-\sin x=0\) at \(\pi/4\) and \(5\pi/4\). Since \(f''=-\sin x-\cos x\), the local maximum is \((\pi/4,\sqrt2)\), and the local minimum is \((5\pi/4,-\sqrt2)\).
Example 7: Interpretation in context. A profit model is \(P(q)=-q^3+12q^2-36q+100\) for \(q>0\). Since \(P'(q)=-3(q-2)(q-6)\), the stationary quantities are \(2\) and \(6\). Also \(P''(q)=-6q+24\), so \(P''(2)>0\) gives a local minimum profit of \(P(2)=68\), while \(P''(6)<0\) gives a local maximum profit of \(P(6)=100\).
7. AP Reasoning Routine
State theorem hypotheses, make sign charts on domain intervals, include endpoints when required, and connect derivative signs to function behavior.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Use the Second Derivative Test where possible. If it is inconclusive or unavailable, say so and use an appropriate alternative.
(a) Classify all local extrema of \(f(x)=x^3-6x^2+9x\).
(b) Classify all local extrema of \(f(x)=x^4-8x^2\).
(c) Analyze every critical number of \(f(x)=x^4+4x^3\).
(d) Compare the test's result at zero for \(f(x)=x^4\) and \(g(x)=-x^4\), then finish the classifications.
(e) Apply the test to \(f(x)=x^3\) at zero and resolve the result.
(f) Find and classify the stationary point of \(f(x)=e^x-3x\).
(g) Classify the stationary points of \(f(x)=\cos x\) on \((0,2\pi)\).
(h) Suppose \(f'(a)=f'(b)=f'(d)=0\), with \(f''(a)=5\), \(f''(b)=-2\), and \(f''(d)=0\). State every justified conclusion.
(i) Explain what to do when \(f'(c)=0\) but \(f''(c)\) is undefined.
(j) Explain why a local minimum established by the Second Derivative Test need not be the absolute minimum on a closed interval.
Check the solution
In part (a), \(f'(x)=3(x-1)(x-3)\) and \(f''(x)=6x-12\). Since \(f''(1)=-6\), \((1,4)\) is a local maximum; since \(f''(3)=6\), \((3,0)\) is a local minimum. In part (b), \(f'(x)=4x(x-2)(x+2)\) and \(f''(x)=12x^2-16\). Thus \((0,0)\) is a local maximum and \((-2,-16)\) and \((2,-16)\) are local minima. In part (c), \(f'(x)=4x^2(x+3)\), so the critical numbers are \(-3\) and \(0\). Since \(f''(x)=12x(x+2)\), \(f''(-3)=36>0\), giving a local minimum at \((-3,-27)\). At zero the second derivative is zero, so the test is inconclusive; \(f'\) is positive on both sides of zero, making it neither a local maximum nor a local minimum. In part (d), the test is inconclusive for both functions because both second derivatives equal zero at zero. The First Derivative Test or direct comparison shows a local minimum for \(x^4\) and a local maximum for \(-x^4\). In part (e), \(f'(0)=f''(0)=0\), so the test is inconclusive. Because \(f'(x)=3x^2>0\) on both sides, zero is neither type. In part (f), \(f'(x)=e^x-3=0\) at \(x=\ln3\). Since \(f''(\ln3)=3>0\), the local minimum is \((\ln3,3-3\ln3)\). In part (g), \(f'(x)=-\sin x\) vanishes at \(x=\pi\) inside the interval. Since \(f''(\pi)=1>0\), \((\pi,-1)\) is a local minimum. In part (h), \(a\) gives a local minimum, \(b\) gives a local maximum, and the test is inconclusive at \(d\). In part (i), the Second Derivative Test cannot classify the point; analyze the sign of \(f'\) on the two sides or use another local argument. In part (j), the test compares behavior only near the stationary point. Absolute extrema require comparison with all other critical candidates and included endpoints.