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Second Derivative, Concavity and Inflection Points
Key Terms
- The second derivative f″(x) = d²y/dx² measures how the gradient is changing
- f″(x) > 0 ⇒ curve is concave up (holds water, gradient is increasing)
- f″(x) < 0 ⇒ curve is concave down (sheds water, gradient is decreasing)
- An inflection point is where concavity changes — i.e. f″(x) changes sign
- f″(a) = 0 is necessary but NOT sufficient: you must confirm a sign change of f″ around x = a
f″(x) > 0 on interval ⇒ concave up on that interval
f″(x) < 0 on interval ⇒ concave down on that interval
Finding Inflection Points:
1. Find f″(x)
2. Solve f″(x) = 0 (and check where f″ is undefined)
3. Test the sign of f″(x) on each side of each candidate
4. If the sign changes, the point is an inflection point
Notation: f″(x) = d²y/dx² = (f′)′(x)
Step 1: f′(x) = 4x³ − 12x²
Step 2: f″(x) = 12x² − 24x = 12x(x − 2)
Step 3: f″(x) = 0 ⇒ x = 0 or x = 2
Step 4 — Sign test:
x < 0: f″(−1) = 12(1)(3) = 36 > 0 (concave up)
0 < x < 2: f″(1) = 12(1)(−1) = −12 < 0 (concave down)
x > 2: f″(3) = 12(9)(1) = 108 > 0 (concave up)
Sign changes at both x = 0 and x = 2.
f(0) = 0, f(2) = 16 − 32 = −16
Inflection points: (0, 0) and (2, −16)
f′(x) = ex − 3
f″(x) = ex
Since ex > 0 for all x, f″(x) > 0 everywhere.
Conclusion: f is concave up on all of (−∞, +∞) and has no inflection points.
Understanding the Second Derivative
The first derivative f′(x) tells us the gradient (slope) of a curve at any point — whether the function is increasing or decreasing. The second derivative f″(x) goes one level deeper: it measures the rate of change of the gradient. In other words, it tells us whether the gradient itself is increasing or decreasing as we move along the curve.
Think of driving a car. Your position is like f(x), your speed (rate of change of position) is like f′(x), and your acceleration (rate of change of speed) is like f″(x). Positive acceleration means you are speeding up; negative acceleration means you are slowing down. The second derivative plays exactly the same role for the gradient of a curve.
Formally, f″(x) = d/dx[f′(x)]. We differentiate f once to get f′, then differentiate again to get f″. This is also written as d²y/dx² in Leibniz notation, reinforcing the idea of differentiating twice.
Concavity: What It Means Geometrically
A curve is concave up on an interval when it curves upward like a bowl or the letter U. On a concave-up section, the gradient is increasing — the curve gets steeper and steeper as you move right (or less steep and then steeper again after a turning point). Mathematically, concave up corresponds to f″(x) > 0.
A curve is concave down on an interval when it curves downward like an arch or an upside-down bowl. On a concave-down section, the gradient is decreasing. Mathematically, concave down corresponds to f″(x) < 0.
A useful memory device: when f″(x) > 0, the curve holds water (concave up); when f″(x) < 0, the curve sheds water (concave down). Another way to think about it: on a concave-up section, a tangent line drawn at any point sits below the curve; on a concave-down section, the tangent line sits above the curve.
Inflection Points: Where Concavity Changes
An inflection point is a point on the curve where the concavity changes — either from concave up to concave down, or from concave down to concave up. At an inflection point, the curve crosses its own tangent line. Visually, the curve stops bending one way and starts bending the other.
To find inflection points algebraically, we look for values of x where f″(x) = 0 or where f″(x) is undefined. However, finding these candidates is only the first step. We must then test the sign of f″(x) on both sides of each candidate point. Only if the sign genuinely changes (positive to negative, or negative to positive) do we have a true inflection point.
This sign-change requirement is critical. Consider f(x) = x4: we have f″(x) = 12x², so f″(0) = 0. But for all x ≠ 0, f″(x) = 12x² > 0 — the sign does NOT change. The concavity remains upward on both sides of x = 0, so (0, 0) is NOT an inflection point. Failing to perform this check is one of the most common errors in the QLD exam.
Procedure for Finding Inflection Points
Step 1: Differentiate f(x) twice to find f″(x).
Step 2: Solve f″(x) = 0. Also identify any x-values where f″(x) does not exist (these are additional candidates).
Step 3: For each candidate x = a, choose a test point to the left and a test point to the right. Evaluate the sign of f″ at each test point.
Step 4: If the sign changes, x = a is an inflection point. Calculate f(a) to get the coordinates.
Step 5: State your answer with both coordinates — examiners expect the full point (a, f(a)), not just the x-value.
Describing Concavity on Intervals
Exam questions often ask you to describe where a function is concave up or concave down. After finding the inflection points (where concavity changes), you know the curve alternates between concave up and concave down on the intervals between these points. Test one value in each interval to determine which type of concavity applies there.
For example, if inflection points occur at x = −1 and x = 3, test a value in (−∞, −1), in (−1, 3), and in (3, +∞). The sign of f″ in each interval tells you the concavity there.
Connecting Concavity to Curve Sketching
Knowledge of concavity gives you vital information for sketching curves accurately. Stationary points (where f′ = 0) can be classified as local maxima (concave down at that point) or local minima (concave up at that point) using the sign of f″ — this is called the second derivative test, which will be covered in the next lesson. More broadly, correctly sketching concavity — showing where the curve bends upward vs downward and where inflection points occur — is a key part of any complete curve sketch.
Mastery Practice
- Fluency Find f″(x) for f(x) = 3x4 − 5x³ + 2x − 7.
- Fluency Find f″(x) for f(x) = 2ex + 5 sin x − 3x².
- Fluency For f(x) = x³ − 6x, determine whether the curve is concave up or concave down at x = 2 and at x = −1.
- Fluency For f(x) = 4 cos x, find f″(x) and describe the concavity when 0 < x < π.
- Understanding Find all inflection points of f(x) = x³ − 3x² − 9x + 2.
- Understanding Find all inflection points of f(x) = x4 − 4x³ + 6x².
- Understanding Find the intervals of concavity and inflection points for f(x) = 2x3 + 3x2 − 12x.
- Understanding A student claims that f(x) = x4 has an inflection point at x = 0 because f″(0) = 0. Is this correct? Justify your answer fully.
- Problem Solving The position of an object is given by s(t) = t4 − 8t3 + 18t2 for t ≥ 0.
(a) Find the times when the acceleration is zero.
(b) Determine whether there is a change in the nature of acceleration at each of these times. - Problem Solving For f(x) = ex − 4x:
(a) Show that f has no inflection points.
(b) Determine the single interval on which f is concave up.
(c) Explain what the concavity tells us about the shape of the graph.