# Higher Derivatives and Concavity

Mathematics I · Differential Calculus · https://tryals.app/learn/mathematics-i/higher-derivatives-and-concavity

## The Second Derivative

Differentiating a derivative gives the **second derivative** $f''(x)$, the rate of change of the rate of change. If $f$ is position, $f'$ is velocity and $f''$ is **acceleration**. Higher derivatives ($f'''$, etc.) continue this pattern.

The second derivative reveals **concavity**, how a curve bends:

| Sign | Concavity | Visual | Slope |
| --- | --- | --- | --- |
| $f'' > 0$ | **Concave up** | Cup-shaped | **Increasing** |
| $f'' < 0$ | **Concave down** | Cap-shaped | **Decreasing** |

A point where concavity changes sign is an **inflection point**. While $f''(x) = 0$ is necessary, it is **not sufficient**: concavity must actually switch signs.

## Tests and Curve Sketching

Concavity gives the **second-derivative test** for critical points where $f'(x) = 0$:

| Condition | Classification | Visual |
| --- | --- | --- |
| $f'' > 0$ | **Local minimum** | Bottom of cup |
| $f'' < 0$ | **Local maximum** | Top of cap |
| $f'' = 0$ | **Inconclusive** | Use 1st derivative |

> **Pitfall:** $f''(x) = 0$ does **not** guarantee an inflection point. For $f(x) = x^4$, $f''(0) = 0$, but it is concave up everywhere, so $x=0$ is **not** an inflection point.

Together, both derivatives determine shape, driving **curve sketching** and optimization.

## Practice questions

6 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. A cyclist's speed is falling and their acceleration is negative, but a news report says the *rate of decrease* is slowing. Which derivative is that, and what is happening?

A. The statement is contradictory: a decreasing quantity cannot decrease more slowly
B. The first derivative, which is negative and becoming less negative
C. The third derivative, since it describes a change in acceleration
D. The second derivative, since acceleration is easing toward zero

**Answer:** D. The second derivative, since acceleration is easing toward zero

**Why:** Acceleration is negative and heading toward zero, so the second derivative is positive: braking that is easing off. Nested rate-of-change language is exactly where intuition fails, and it is the same structure as the epidemic reported as 'cases still rising, but more slowly', which is a claim about the second derivative.

Page: https://tryals.app/practice/mathematics-i/higher-derivatives-and-concavity/a-cyclists-speed-is-falling-and-their-acceleration-is-negative-but

### 2. Let c be a critical point of f, so f'(c) = 0. Select every correct conclusion.

A. If f''(c) = 0 the second-derivative test gives no verdict
B. f''(c) > 0 means the curve is concave up at c
C. If f''(c) > 0 then c is a local minimum
D. If f''(c) < 0 then c is a local maximum
E. A vanishing second derivative is enough to guarantee an inflection

**Answer:** A. If f''(c) = 0 the second-derivative test gives no verdict; B. f''(c) > 0 means the curve is concave up at c; C. If f''(c) > 0 then c is a local minimum; D. If f''(c) < 0 then c is a local maximum

**Why:** Concave up at a flat point means the bottom of a cup, hence a minimum, and concave down gives a maximum. When f'' = 0 the test is silent: x to the fourth has a minimum there, x cubed has neither, and minus x to the fourth has a maximum, all three with f'' = 0.

Page: https://tryals.app/practice/mathematics-i/higher-derivatives-and-concavity/let-c-be-a-critical-point-of-f-so-f-c-0-select-every-correct

### 3. Match each derivative condition to what it says about the graph.

**Answer:**

- f' is positive → The function is rising
- f'' is positive → The curve is concave up
- f'' is negative → The curve is concave down
- f'' changes sign → There is an inflection point

**Why:** Direction comes from f' and bending from f''; the two are independent, so a curve can rise while bending either way. Only the last entry involves a change of sign, which is why inflection points cannot be detected from a single value of f''.

Page: https://tryals.app/practice/mathematics-i/higher-derivatives-and-concavity/match-each-derivative-condition-to-what-it-says-about-the-graph

### 4. Order the steps of locating the inflection points of a twice-differentiable function.

**Answer:**

1. Compute the second derivative
2. Solve f'' (x) = 0 and note any points where f'' is undefined
3. Test the sign of f'' on each side of every candidate
4. Keep only the candidates where the sign actually changes

**Why:** The last two steps are the ones that get skipped, and skipping them wrongly promotes x = 0 on the graph of x to the fourth. Candidates must be filtered by an actual sign change, which is the definition of an inflection point.

Page: https://tryals.app/practice/mathematics-i/higher-derivatives-and-concavity/order-the-steps-of-locating-the-inflection-points-of-a

### 5. If f''(c) = 0, the point c need not be an inflection point - x⁴ at x = 0 is the standard counterexample. Why is a vanishing second derivative not enough?

A. Because concavity must change sign, and for x⁴ it does not.
B. Because inflection points require the first derivative to be zero as well, which fails here.
C. Because the second derivative must be undefined rather than zero at an inflection point.
D. Because x⁴ is not differentiable twice at the origin, so the test does not apply.

**Answer:** A. Because concavity must change sign, and for x⁴ it does not.

**Why:** For x⁴ the second derivative is 12x², zero at the origin and positive on both sides, so the curve never changes its bending. An inflection needs a sign change, as x³ supplies. It is exactly parallel to the first-derivative case, f'(c) = 0 does not give an extremum, since x³ has a horizontal tangent and neither. Zeros locate candidates; sign changes confirm them.

Page: https://tryals.app/practice/mathematics-i/higher-derivatives-and-concavity/if-f-c-0-the-point-c-need-not-be-an-inflection-point-x-at-x

### 6. Sort each function by whether x = 0 is an inflection point of its graph.

**Answer:**

- x = 0 is an inflection point: x cubed, x to the fifth, x cubed minus x
- x = 0 is not an inflection point: x to the fourth, x squared, x to the fourth plus x

**Why:** For the even powers f'' is 12x squared or 2, neither of which changes sign, so those graphs stay concave up on both sides. The odd powers give f'' = 6x and 20x cubed, which do switch sign. Adding the linear term x changes the slope but not f'' at all, so item 5 still fails.

Page: https://tryals.app/practice/mathematics-i/higher-derivatives-and-concavity/sort-each-function-by-whether-x-0-is-an-inflection-point-of-its