5.7 Using the Second Derivative Test to Determine Extrema
You’ve probably noticed by now that Unit 5 deals with analytical applications of differentiation; that means that a function’s derivatives can tell us something about its behaviors. We learned from 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema that the first derivative of a function tells us information on where its relative extrema (aka local maximum and minimum) will be.
What does the second derivative tell us, then? Answer: Whether a critical point we found yields a local maximum or minimum (it can’t be both)! 🧠
✏️ Warm-up: Finding Critical Points
Let’s do a quick recap of critical points from key topic 5.2! To refresh, a critical point is a point on a function where…
- the first derivative equals zero or
- fails to exist
Analytically, the first bullet point is easier to wrap your head around. Let’s take a look at a quick example:
Computing the first derivative gives us:
To find where the first derivative equals zero, we can factor and then set :
Looks familiar, right? We proceed as if we’re finding the roots (x) in a regular equation. In this case, the two x values are our critical points!
Not bad, right? What do we do with these critical point values, then?
We learned two things from 5.6 Determining Concavity of Functions over Their Domains about what the second derivative of a function can tell us information of:
- 🌈 The function’s intervals of upward or downward concavity.
- is positive (or greater than zero, aka ) = concave up
- is negative (or less than zero, aka ) = concave down
- 🪝 The function graph’s inflection points (where concavity changes).
🥈 Extending the Second Derivative Test
Now, let’s connect these ideas to the critical points we mentioned earlier: by knowing the concavities before and after the critical points, we can determine where our local minima and maxima are! 🗺️
🪜 Second Derivative Test Steps
Here are some steps that we’ll go through:
- Find the critical points of using .
- Plug the critical points into to get a y-value.
- Determine whether you have a local min or max!
💡 Key Idea:
- If the y-value from Step 2 is positive, you have a local minimum! 🥣
- If the y-value from Step 2 is negative, you have a local maximum! 🗻
📝 Second Derivative Test Walkthrough
To illustrate, let’s continue with our example while we walk through the steps pertaining to the Second Derivative Test.
🥇 Step 1: Find ’s critical points using
Done! We found our critical points: and earlier.
🥈 Step 2: Plug critical points into to get a y-value.
From above, we found to be:
Calculating the second derivative:
Then, we plug our critical points:
These numbers give us the y-values for our local extrema and thus the whole coordinates: and . We’re ready for our final step…
🥉 Step 3: Determine whether you have a local min or max!
In our case, is a local maximum and is a local minimum. This is because of the work we did in step 2!
Conceptually, why does this make sense? Tap into your imagination… imagine a hill and a bowl!
- 🗻 If a point represents a maximum and is situated at the peak of a hill, the point appears to lie on the concave-down surface of said hill.
- 🥣 Conversely, if a point signifies a minimum and is located at the bottom of an upright bowl, it suggests that the point lies on the concave-up surface of the bowl.
Therefore, when the function f is concave down at a critical point (we’ll call it “c”), it implies a maximum, whereas f being concave up at “c” indicates a minimum.
Feel more confident? Try the following examples on your own before looking at the solutions to see which areas you should do more practice on! 📣
💭 Second Derivative Test: Let’s Practice!
For each of the following functions f(x), determine whether each of their critical points are local maxima or minima.
Second Derivative Test: Example 1
Let’s find our first and second derivatives:
Setting to find critical points x:
Thinking back to the Unit Circle from trigonometry, at which angles $x$ will $cos(x) = 0$? We’ll get our critical points:
Plugging these values into our second derivative:
Since , is a local maximum of . On the other hand, since , is a local minimum of f(x). We’re done! ⭐
Second Derivative Test: Example 2
Calculating f’(x) and f’’(x) gives us:
Setting …
Wait a second. This doesn’t give us a solution. What do we do in this case, then? This brings up another element of the Second Derivative Test:
💡 If and , or if doesn't exist, then the test is inconclusive. Inconclusive, in this context, means that there may be a point of inflection; this may also mean that there might be a local maximum or minimum that we can’t pinpoint exactly.
💫 Closing
Great work! Now we applied the second derivative test to different functions to determine extrema. There’s one key point that we want to leave you with…
When dealing with a continuous function that has just one critical point across its entire domain and this particular point serves as a local extremum within a specific interval, it also stands as the global extremum for the function within that particular interval.
This is because there are no other critical points to challenge its extremum status within that specific interval. This is due to the function having only one peak or valley (extremum) within that interval. 🗻
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| absolute maximum | The highest value of a function over its entire domain or a specified interval. |
| absolute minimum | The lowest value of a function over its entire domain or a specified interval. |
| continuous | A function that has no breaks, jumps, or holes in its graph over a given interval. |
| critical point | A point in the domain of a function where the derivative is zero or undefined, which are candidates for local and absolute extrema. |
| global extremum | The absolute maximum or minimum value of a function over its entire domain or a specified interval. |
| relative maximum | A point where a function reaches a highest value in a neighborhood around that point. |
| relative minimum | A point where a function reaches a lowest value in a neighborhood around that point. |
| second derivative | The derivative of the first derivative, denoted f'', which describes the concavity of a function and indicates where it is concave up or concave down. |
Frequently Asked Questions
How do I use the second derivative test to find max and min points?
Start by finding critical points: solve f′(x)=0 or where f′ is undefined. For each critical point c do the Second Derivative Test: - Compute f′′(c). - If f′′(c) > 0, f is concave up at c → local minimum at c. - If f′′(c) < 0, f is concave down at c → local maximum at c. - If f′′(c) = 0 or undefined, the test is inconclusive—use the First Derivative Test (check sign changes of f′ around c) or analyze higher derivatives/graph. Remember endpoints: on a closed interval check endpoints too for absolute extrema (Candidates Test). Also note FUN-4.A.8: if a continuous function has only one critical point on an interval and that critical point is a local extremum, it’s also the absolute extremum on that interval. This is exactly what AP asks you to justify about derivatives and concavity (Topic 5.7, FUN-4.A). For a quick review see the topic study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep), the Unit 5 overview (https://library.fiveable.me/ap-calculus/unit-5), and practice problems (https://library.fiveable.me/practice/ap-calculus).
What's the formula for the second derivative test?
If c is a critical point with f′(c) = 0 and f is twice differentiable near c, the Second Derivative Test says: - If f″(c) > 0, f is concave up at c ⇒ c is a local (relative) minimum. - If f″(c) < 0, f is concave down at c ⇒ c is a local (relative) maximum. - If f″(c) = 0 (or f″ is undefined), the test is inconclusive—use the First Derivative Test, higher derivatives, or analyze sign changes of f′. Remember: this test only applies at interior critical points where f′(c)=0 and you know f″(c). Endpoints and points where f′ is undefined must be checked separately (Candidates Test/endpoint analysis) to find absolute extrema. This aligns with the CED keywords: Critical Point, Concavity, Local/Absolute Extremum, and when f″ = 0 is inconclusive. For a quick review, see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep) and practice problems (https://library.fiveable.me/practice/ap-calculus).
When do I use the second derivative test vs the first derivative test?
Use the Second Derivative Test when you have a critical point c (f′(c)=0) and you can compute f″(c) easily: if f″(c)>0 then f has a local min at c (concave up), if f″(c)<0 then a local max (concave down)—this is exactly FUN-4.A.7 in the CED. But don’t use it if f″(c)=0 or f″ is undefined at c; the test is inconclusive then. Use the First Derivative Test whenever you want a foolproof method: check the sign of f′ to the left and right of c. If f′ changes +→− it’s a local max, −→+ a local min, no sign change means no extremum. The first derivative test always works (and handles cases where f″(c)=0 or is hard to compute). Don’t forget endpoints and the Candidates Test for absolute extrema (compare critical points and endpoints)—see Topic 5.5 and 5.4 (unit overview: https://library.fiveable.me/ap-calculus/unit-5; second-derivative guide: https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep). For extra practice, use the AP practice problems (https://library.fiveable.me/practice/ap-calculus).
Can someone explain step by step how to do the second derivative test?
Step-by-step second-derivative test (quick and AP-safe): 1. Find critical points: compute f′(x) and solve f′(c) = 0 or where f′ is undefined. (These are candidates for local extrema per CED: critical point / stationary point.) 2. Compute f′′(x) and evaluate at each critical c (if f′′ exists at c). 3. Apply the test: - If f′′(c) > 0 → f is concave up at c → local minimum at c. - If f′′(c) < 0 → f is concave down at c → local maximum at c. - If f′′(c) = 0 or f′′(c) is undefined → test is inconclusive. Use the first-derivative test or examine higher derivatives / graph shape (inflection points possible). 4. Don’t forget endpoints and global extrema: on a closed interval, evaluate f at critical points AND endpoints; the largest/smallest value is the absolute (global) max/min. Note FUN-4.A.8: if a continuous function on an interval has only one critical point and it’s a local extremum, it’s also the absolute extremum. AP tip: show work finding f′ and f′′ and justify concavity sign (connects derivatives → behavior). For extra practice, see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep) and more problems at (https://library.fiveable.me/practice/ap-calculus).
What does it mean when f''(x) is positive or negative at a critical point?
If x = c is a critical point (usually f′(c) = 0 or f′ undefined), the sign of f″(c) tells you the concavity there and so often the type of local extremum: - f″(c) > 0: f is concave up at c (cup-shaped). If f′(c) = 0, the Second Derivative Test says c is a local minimum. - f″(c) < 0: f is concave down at c (cap-shaped). If f′(c) = 0, c is a local maximum. Important caveats (AP-style): if f″(c) = 0 or f″ is undefined, the Second Derivative Test is inconclusive—use the First Derivative Test or higher-derivative tests. Also remember to check endpoints and compare values when looking for absolute extrema; if a continuous function has only one critical point on an interval and it’s a local extremum, it’s also the global extremum (FUN-4.A.8). For more examples and AP-aligned practice, see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep) and Unit 5 overview (https://library.fiveable.me/ap-calculus/unit-5).
I'm confused about when the second derivative test fails - what do I do then?
The second-derivative test “fails” whenever f′(c)=0 but f″(c)=0 or f″(c) is undefined—it just doesn’t tell you whether c is a max, min, or neither. When that happens, use the First Derivative Test: check the sign of f′ just left and right of c. If f′ changes +→− you have a local max; −→+ gives a local min; no sign change means no local extremum (stationary/inflection point). You can also: - Check higher derivatives: if the first nonzero derivative after f′ at c is of even order, you get extremum (sign depends); if odd order, no extremum. - Use concavity or graph/limit behavior and include endpoints when finding absolute extrema (Candidates Test). This aligns with CED FUN-4.A (use derivatives to justify behavior). For a quick review see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep) and try practice problems (https://library.fiveable.me/practice/ap-calculus).
How do I find critical points before using the second derivative test?
First find the critical points of f(x)—those are the x-values you test with the Second Derivative Test. 1. Check the domain: only consider x where f is defined (and interior points if you’re looking for local extrema). 2. Compute f′(x). Solve f′(x) = 0 (stationary points). 3. Find where f′(x) is undefined but f(x) is defined (these are also critical points). 4. Don’t forget endpoints of a closed interval—they’re candidates for absolute extrema even though they aren’t “critical” by the derivative test. After you have the critical points, use the Second Derivative Test: if f′(c)=0 and f″(c)>0 → local min; if f″(c)<0 → local max; if f″(c)=0 → inconclusive (use the First Derivative Test or higher derivatives). If there’s only one critical point on a continuous interval and it’s a local extremum, it’s also the absolute extremum (FUN-4.A.8). For AP-aligned review and worked examples, see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep) and more Unit 5 resources (https://library.fiveable.me/ap-calculus/unit-5). Practice lots of problems (https://library.fiveable.me/practice/ap-calculus) to get fast at spotting critical points.
What's the difference between relative extrema and absolute extrema?
A relative (local) extremum is a point where f(x) is higher (local maximum) or lower (local minimum) than all nearby points—formally, f(c) ≥ f(x) (or ≤) for x in some open interval around c. The Second Derivative Test can tell you whether a critical point is a relative max (f′(c)=0 and f″(c)<0) or relative min (f″(c)>0) (CED FUN-4.A.7). An absolute (global) extremum is the largest or smallest value of f on a whole interval: f(c) ≥ f(x) (or ≤) for every x in the interval. To identify absolute extrema you must compare all candidates (critical points and endpoints)—the Candidates Test/endpoint analysis is required on AP problems (CED keywords: Candidates Test, Endpoint Analysis). If a continuous function has only one critical point on an interval and that point is a relative extremum, then it is also the absolute extremum (CED FUN-4.A.8). For more examples, see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep) and Unit 5 overview (https://library.fiveable.me/ap-calculus/unit-5). For extra practice, try problems at (https://library.fiveable.me/practice/ap-calculus).
If there's only one critical point on an interval, is it automatically the absolute max or min?
Not automatically. You need more conditions. Key points from the CED: if a continuous function on an interval has exactly one critical point and that critical point is a local extremum, then it is also the absolute (global) extremum on that interval (FUN-4.A.8). But two things can fail: - The critical point might not be a local max/min (e.g., f(x)=x^3 has the single critical point x=0, but it’s an inflection/stationary point, not an extremum). - The interval endpoints matter on closed intervals: absolute extrema can occur at endpoints, so a unique interior critical point that’s a local extremum will be global only if it’s the only candidate (no endpoint beats it). So rule of thumb: find all critical points and evaluate them and endpoints (Candidates Test). Use the Second Derivative Test to classify a critical point; if it gives a local extremum and it’s the only critical point on a continuous interval, it’s the absolute extremum (see the Topic 5.7 study guide: https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep). For extra practice, try problems at the Unit 5 page (https://library.fiveable.me/ap-calculus/unit-5) or the practice bank (https://library.fiveable.me/practice/ap-calculus).
Why does the second derivative test work for finding concavity?
Think of f′ as the slope of f. The second derivative f″ tells you how that slope is changing. If f″(a) > 0, the slopes to the left and right of a are increasing—the tangent lines tilt upward as x increases—so the graph bends up (concave up) and a critical point (f′(a)=0) is a local minimum. If f″(a) < 0, slopes are decreasing, the graph bends down (concave down), and a critical point is a local maximum. That’s the Second Derivative Test in FUN-4.A: f″’s sign determines concavity, and combining f′(a)=0 with f″(a) ≠ 0 gives the extremum type. If f″(a)=0 or undefined, the test is inconclusive—you must use the First Derivative Test or examine sign changes of f′. For a quick refresher tied to the CED, see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep). For more practice, try problems at (https://library.fiveable.me/practice/ap-calculus).
How do I solve second derivative test problems on the AP exam?
Quick recipe you can use on the exam: 1. Find critical points: compute f′(x)=0 or f′ undefined (only in domain). These are candidates for extrema (CED: Critical Point, FUN-4.A). 2. Compute f′′(x) at each critical point: - If f′′(c) > 0 → f is concave up at c → local minimum. - If f′′(c) < 0 → f is concave down at c → local maximum. - If f′′(c) = 0 or f′′ undefined → second derivative test is inconclusive—use the First Derivative Test or check values around c. (CED: FUN-4.A.7) 3. For absolute (global) extrema on a closed interval, evaluate f at critical points and at endpoints; compare values (CED: FUN-4.A.8). 4. On the AP: show reasoning (justifying concavity from f′′ or sign changes of f′)—free-response wants explanation; multiple-choice expects quicker sign checks. Practice this on the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep) and drill problems (https://library.fiveable.me/practice/ap-calculus).
What happens if the second derivative equals zero at a critical point?
If f''(c) = 0 at a critical point c, the Second Derivative Test is inconclusive—it doesn’t tell you whether c is a local max, local min, or neither. The CED calls this a stationary point that could be an inflection point (like f(x)=x^3: f'(0)=0, f''(0)=0, no extremum) or a higher-order extremum (like f(x)=x^4: f'(0)=0, f''(0)=0, but a local minimum). What to do instead: - Use the First Derivative Test: check how f' changes sign around c. If f' goes +→− you have a local max; −→+ gives a local min; no sign change means not an extremum. - Check f'' on intervals near c to test concavity (concave up/down). - Look at higher derivatives or Taylor expansion: if the first nonzero derivative after f' is of even order and positive → local min; even and negative → local max; odd order → inflection. Also remember to compare endpoints and use the Candidates Test for absolute extrema on closed intervals. For AP review, see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep) and try practice problems (https://library.fiveable.me/practice/ap-calculus).
I don't understand how to tell if a critical point is a max or min using f''(x)
If c is a critical point (usually f′(c) = 0) and f is twice-differentiable near c, use the Second Derivative Test: - If f″(c) > 0, the graph is concave up at c → local (relative) minimum. - If f″(c) < 0, the graph is concave down at c → local (relative) maximum. - If f″(c) = 0 or f″ is undefined, the test is inconclusive—use the First Derivative Test (check sign changes of f′) or higher derivatives. Remember: endpoints of a closed interval can still be absolute (global) extrema even if the second derivative test doesn’t apply. Also note the CED: if a continuous function has only one critical point on an interval and it’s a local extremum, it’s also the absolute extremum (FUN-4.A.7 and FUN-4.A.8). For review and examples, see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep). For more practice, try problems at (https://library.fiveable.me/practice/ap-calculus).
When can I conclude that a local extremum is also a global extremum?
You can only conclude a local extremum is global after you check the domain and any other candidates, not just from the Second Derivative Test alone. Key rules (CED/FUN-4.A): - If f is continuous on a closed interval [a,b], use the Candidates Test: compare f at all critical points and at the endpoints. A local min/max might be the absolute min/max only if its value beats the others. - If f is continuous on an interval and has exactly one critical point in that interval AND that critical point is a local extremum, then it is also the absolute extremum on that interval (FUN-4.A.8). - If the domain is open or unbounded, endpoints may not exist so you must consider limits as x → boundary or ±∞. So: use the Second Derivative Test to identify local extrema, then apply the Candidates Test (include endpoints and limits) or the “single critical point on a continuous interval” fact to decide if it’s global. For more AP-aligned notes, see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep) and Unit 5 review (https://library.fiveable.me/ap-calculus/unit-5). For lots of practice problems, go to https://library.fiveable.me/practice/ap-calculus.
How do I use the second derivative test for word problems about optimization?
For optimization word problems use this routine and the Second Derivative Test to justify your answer (CED keywords: critical point, concavity, local/absolute extremum). 1. Model: write the quantity to optimize as a single function f(x) (include units). 2. Domain: state the feasible interval (constraints, endpoints). 3. Critical points: compute f′(x) and solve f′(x)=0 or where f′ is undefined. 4. Second Derivative Test at each critical point c: - If f″(c) > 0 → f is concave up at c → local minimum. - If f″(c) < 0 → concave down → local maximum. - If f″(c) = 0 → test inconclusive; use first derivative test or examine values. 5. Global (absolute) check: evaluate f at critical points AND at endpoints. If there’s only one critical point on the domain and it’s a local extremum, it’s also the absolute extremum (FUN-4.A.8). 6. State the answer with units and a short justification referencing concavity or endpoint comparisons. This matches AP expectations (use justification, include endpoints). For a quick refresher and examples, see the Topic 5.7 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep). For extra practice problems, try Fiveable’s practice set (https://library.fiveable.me/practice/ap-calculus).