Steady State
MAT161 / Lecture 11
Lecture 11 · Barnett ch. 4 · about 40 min

Second Derivative and Concavity

The derivative of the derivative: what concavity means, how to find inflection points, the second-derivative test as a shortcut for classifying extrema, and diminishing returns as a concavity story.

By the end you can
  • Compute a second derivative and interpret it as the rate of change of the rate of change
  • Determine concave-up and concave-down intervals from the sign of f double-prime
  • Find inflection points, including the point of diminishing returns
  • Apply the second-derivative test to classify critical points

The derivative of the derivative

Just as f′(x)f'(x) measures how ff is changing, f′′(x)f''(x) measures how f′(x)f'(x) is changing: the rate at which the rate of change is itself changing.

Second derivativeFormula sheet →
f′′(x)=ddx[f′(x)]f''(x) = \frac{d}{dx}\big[f'(x)\big]
f′′(x)f''(x)
second derivative of f, read 'f double prime'

Use it when you need to differentiate a function twice, such as to check concavity or apply the second-derivative test.

In economics, if C(x)C(x) is total cost, C′(x)C'(x) is marginal cost, and C′′(x)C''(x) tells you whether marginal cost itself is rising or falling — whether each additional unit is getting more or less expensive to produce than the one before.

Your turn

C(x)=0.01x3−0.6x2+20x+500C(x) = 0.01x^3 - 0.6x^2 + 20x + 500. Find C′′(x)C''(x) at x=20x = 20.

Concavity

ConcavityRulesFormula sheet →
f′′(x)>0 on an interval  ⇒  f is concave up theref′′(x)<0  ⇒  f is concave down theref''(x) \gt 0 \text{ on an interval} \;\Rightarrow\; f \text{ is concave up there} \qquad f''(x) \lt 0 \;\Rightarrow\; f \text{ is concave down there}
concaveupconcave up
graph curves upward, like a cup
concavedownconcave down
graph curves downward, like a frown

Use it when you need to describe the curvature of a graph, or confirm whether a critical point is a max or a min via the second-derivative test.

Check your understanding

A production function Q(L)Q(L) satisfies Q′′(L)<0Q''(L) \lt 0 for all L>0L \gt 0. What does this describe?

Inflection points

An inflection point is where concavity switches, from up to down or down to up. Just as critical numbers are candidates for extrema, points where f′′(x)=0f''(x) = 0 (or is undefined) are candidates for inflection points — and just as before, you must check that concavity actually changes.

Inflection pointDefinitionFormula sheet →
f′′(c)=0 (or undefined), and f′′ changes sign at cf''(c) = 0 \text{ (or undefined)}, \text{ and } f'' \text{ changes sign at } c
cc
a candidate inflection point, whose sign change must still be confirmed, exactly as with the first-derivative test

Use it when a question asks for the point of diminishing returns, or any point where a curve changes from bending one way to the other.

Worked example · Finding the point of diminishing returns0/5

Output is Q(L)=−0.02L3+3L2Q(L) = -0.02L^3 + 3L^2, for L>0L \gt 0. Find the point of diminishing returns: where the marginal product Q′(L)Q'(L) stops rising and starts falling.

Your turn

For f(x)=x3−6x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1, find the x-coordinate of the inflection point.

The second-derivative test

Rather than building a full sign chart of f′f', you can often classify a critical point faster by checking the sign of f′′f'' at that single point.

Second-derivative testKey ruleFormula sheet →
f′(c)=0 and f′′(c)>0  ⇒  local min at cf′(c)=0 and f′′(c)<0  ⇒  local max at cf'(c) = 0 \text{ and } f''(c) \gt 0 \;\Rightarrow\; \text{local min at } c \qquad f'(c)=0 \text{ and } f''(c) \lt 0 \;\Rightarrow\; \text{local max at } c
iff′′(c)if f''(c)
0, the test is inconclusive and the first-derivative test must be used instead

Use it when you already have f'' available (or it is easy to compute) and just need to classify a single critical point, rather than build a whole sign chart.

Worked example · Second-derivative test in action0/4

Profit is P(x)=−x3+15x2−48x+200P(x) = -x^3 + 15x^2 - 48x + 200, with critical numbers at x=2x=2 and x=8x=8 (from the previous lesson). Classify both using the second-derivative test.

Your turn

f(x)=x3−12x+7f(x) = x^3 - 12x + 7. Use the second-derivative test to find f''(2).

Exam practice

Exam question 1

f(x)=2x3−9x2+12xf(x) = 2x^3 - 9x^2 + 12x. Find f''(x) at x = 3.

Exam question 2

A cost function has C′′(x)>0C''(x) \gt 0 for all x. What does this say about marginal cost?

Exam question 3

f(x)=x3−15x2+6x−2f(x) = x^3 - 15x^2 + 6x - 2. Find the x-coordinate of the inflection point.

Exam question 4

At a critical point, f′(c)=0f'(c) = 0 and f′′(c)=0f''(c) = 0. What should you do?

Summary and review

Review deck · 9 cards0/9 mastered