Higher Derivatives, Concavity, and Absolute Extrema

Higher-Order Derivatives

• “Higher” means any derivative beyond the first; the most common is the second derivative.
• Standard notations
– f′′(x)f''(x) (two prime marks)
– d2ydx2\dfrac{d^2y}{dx^2}
– D<em>x  2[f]D<em>x^{\;2}[f] or, more compact, d</em>x  2(f)d</em>x^{\;2}(f)
• Third derivative: f′′′(x)f'''(x). After the third, we switch to parenthetical superscripts: f(n)(x)f^{(n)}(x), n≥4n\ge 4.
– Warning: A parenthetical superscript denotes an n-th derivative, not an exponent of the function.
• For a polynomial of degree nn we can take n+1n+1 derivatives before obtaining 00 (e.g.
a degree-4 polynomial yields a 5-th derivative of 00).

Quick Polynomial Example
  1. Given f(x)=5x4−4x3+3x+7f(x)=5x^4-4x^3+3x+7
  2. f′(x)=20x3−12x2+3f'(x)=20x^3-12x^2+3
  3. f′′(x)=60x2−24xf''(x)=60x^2-24x
  4. Evaluate at x=1!:  f′′(1)=60−24=36x=1!:\;f''(1)=60-24=36.

Worked Second-Derivative Examples

(1) Chain Rule + Product Rule Combo

• f(x)=2 (x3+1)2f(x)=2\,(x^3+1)^2
1st derivative (chain rule):
f′(x)=2⋅2(x3+1)⋅3x2=6x2 (x3+1)f'(x)=2\cdot 2(x^3+1)\cdot 3x^2=6x^2\,(x^3+1)
Rewrite before differentiating again:
f′(x)=6x2 (x3+1)f'(x)=6x^2\,(x^3+1)
2nd derivative (product rule):
f′′(x)=(12x)(x3+1)+6x2 (3x2)=12x(x3+1)+18x4f''(x)=\bigl(12x\bigr)(x^3+1)+6x^2\,(3x^2)=12x(x^3+1)+18x^4
f′′(x)=30x4+12x\boxed{f''(x)=30x^4+12x}

(2) Pure Product Rule with xexx e^{x}
  1. f′(x)=1⋅ex+xex=ex(1+x)f'(x)=1\cdot e^{x}+x e^{x}=e^{x}(1+x)
  2. f′′(x)=ex(1+x)+ex=ex(2+x)f''(x)=e^{x}(1+x)+e^{x}=e^{x}(2+x)
  3. Factorised form ex(x+2)e^{x}(x+2) is desirable for the second-derivative test.
(3) Quotient Rule with ln⁡xx\tfrac{\ln x}{x}

• f(x)=ln⁡xxf(x)=\dfrac{\ln x}{x} (domain x>0)

1st derivative (quotient rule):
f′(x)=(1/x) x−ln⁡x⋅1x2=1−ln⁡xx2f'(x)=\dfrac{(1/x)\,x-\ln x\cdot 1}{x^{2}}=\dfrac{1-\ln x}{x^{2}}

2nd derivative (quotient rule again):
Let g(x)=x2g(x)=x^{2}, g′(x)=2xg'(x)=2x, h(x)=1−ln⁡xh(x)=1-\ln x, h′(x)=−1/xh'(x)=-1/x
f′′(x)=g h′−h g′g2=x2 (−1/x)−(1−ln⁡x)(2x)x4f''(x)=\frac{g\,h' - h\,g'}{g^{2}}=\frac{x^{2}\,(-1/x)-(1-\ln x)(2x)}{x^{4}}
=−x−2x+2xln⁡xx4=2ln⁡x−3x3=\frac{-x-2x+2x\ln x}{x^{4}}=\frac{2\ln x-3}{x^{3}}
f′′(x)=2ln⁡x−3x3\boxed{f''(x)=\dfrac{2\ln x-3}{x^{3}}}

Physical Interpretation: Position → Velocity → Acceleration

• Given position s(t)=t3−6t2+9ts(t)=t^{3}-6t^{2}+9t (feet, tt in seconds)
– Velocity v(t)=s′(t)=3t2−12t+9v(t)=s'(t)=3t^{2}-12t+9
– Acceleration a(t)=v′(t)=6t−12a(t)=v'(t)=6t-12

Velocity sign analysis → forward/backward motion:
• Solve v(t)=0⇒t=−2,4v(t)=0\Rightarrow t=-2,4, discard t<0t<0. • Sign chart shows: – 0<t<40<t<4: v<0v<0 ⇒ moving backward. – t>4t>4: v>0 ⇒ moving forward.

Acceleration sign analysis → speeding up/slowing down:
• a(t)=0a(t)=0 at t=1t=1.
• 0<t<10<t<1: a<0a<0 ⇒ slowing down. • t>1t>1: a>0 ⇒ speeding up.

Concavity & Inflection Points

• Concave up (cup) ↔ f''(x)>0 on an interval.
• Concave down (cap) ↔ f''(x)<0 on an interval.
• Inflection point: a point where
– f′′(c)=0f''(c)=0 or does not exist, and
– concavity changes sign across x=cx=c.

Visual cues:
• Concave up: slopes (+→0→-) or increasing slopes throughout.
• Concave down: slopes (-→0→+) or decreasing slopes throughout.
• “Hold water / spill water” analogy.

Testing for Concavity (Concavity Test)
  1. Compute f′′(x)f''(x).
  2. Solve f′′(x)=0f''(x)=0 and find where f′′f'' DNE ⇒ candidate inflection points.
  3. Create intervals; pick test points; plug into f′′f''.
  4. ++ ⇒ concave up; −- ⇒ concave down.

Example (factored f′′f''):
f′′(x)=20x(x−3)(x+3)f''(x)=20x(x-3)(x+3)
Zeros at x=−3,0,3x=-3,0,3 → intervals show
– (−∞,−3)(-\infty,-3): down
– (−3,0)(-3,0): up
– (0,3)(0,3): down
– (3,∞)(3,\infty): up

Second-Derivative Test for Relative Extrema

Prerequisites:
• f′(c)=0f'(c)=0 (critical number) and f′′f'' exists near cc.

Rules:
• If f''(c)>0 → ff has a local minimum at x=cx=c (curve is concave up).
• If f''(c)<0 → ff has a local maximum at x=cx=c (concave down).
• If f′′(c)=0f''(c)=0 or DNE → test is inconclusive; revert to first-derivative test.

Example:
• f′(x)=−6(x+3)(x−4)f'(x)=-6(x+3)(x-4) ⇒ critical numbers x=−3,4x=-3,4.
• f′′(x)=−12x+6f''(x)=-12x+6
– f''(-3)=42>0 ⇒ local min.
– f''(4)=-42<0 ⇒ local max.
• The function also has an inflection point at x=12x=\tfrac12 (where f′′=0f''=0 and concavity changes sign).

Diminishing Returns & “Sweet Spot”

• The inflection point on a revenue or output curve pinpoints when increasing inputs stop increasing outputs at an increasing rate.
• To maximise return per unit input, operate at or just before the inflection point.

Absolute (Global) Extrema

Definitions

• f(c)f(c) is an absolute maximum on interval II if f(c)≥f(x)f(c)\ge f(x) for every x∈Ix\in I.
• f(c)f(c) is an absolute minimum on II if f(c)≤f(x)f(c)\le f(x) for all x∈Ix\in I.

Extreme Value Theorem (EVT)

• A continuous function on a closed interval [a,b][a,b] attains both an absolute max and min.

Critical Point Theorem (single-critical-point version)

• If ff is continuous on ((a,b),hasexactly<strong>one</strong>criticalnumber, has exactly <strong>one</strong> critical numberc in that interval, then
– If the point is a relative max ⇒ it is the absolute max.
– If the point is a relative min ⇒ it is the absolute min.

Checklist for Finding Absolute Extrema on a Closed Interval
  1. Compute f'(x);findallcriticalnumbersin; find all critical numbers in(a,b).
  2. Evaluate fateachcriticalnumber<strong>and</strong>attheendpointsat each critical number <strong>and</strong> at the endpointsa,b.
  3. Largest value ⇒ absolute max; smallest ⇒ absolute min.
If Interval Is Unbounded (all real x)

• Replace “endpoints” with end behaviour: evaluate limits as x\to\pm\infty.
• For polynomials:
– Even degree & positive leading coeff ⇒ both ends \to +\infty, so no absolute max; absolute min possible.
– Even degree & negative leading coeff ⇒ both ends \to -\infty, so no absolute min; absolute max possible.
– Odd degree ⇒ ends opposite ⇒ no absolute extrema unless restricted.

Example 1 (closed interval)

Function (given graphically): candidates x=-1,0,2,8. Evaluations:
– f(-1)=-15
– f(0)=-19.05
– f(2)=0
– f(8)=192
Absolute min =-19.05atatx=0;absolutemax; absolute max=192atatx=8.

Example 2 (with undefined derivative at an endpoint)

For f(x)=x^{2/3}-5x^{1/3}+2onon[0,8]:

  1. f'(x)=\dfrac{2}{3}x^{-1/3}-\dfrac{5}{3}x^{-2/3}
    – Critical numbers: x=0(derivativeundefinedbutindomain);(derivative undefined but in domain);x=\tfrac25 (sets numerator to 0).
  2. Evaluate f(0)=2,,f(2/5)≈0.977,,f(8)≈-84.
  3. Absolute max ≈2atatx=0;absolutemin; absolute min≈-84atatx=8.
Example 3 (unbounded interval, even degree)

f(x)=-x^{4}+4x^{3}+20,domain, domain(-\infty,\infty).
• Critical numbers: x=-4,0,1;values; values148,20,23.
• End behaviour: both ends \to -\infty ⇒ no absolute min.
• Largest finite value =148⇒absolutemaxat⇒ absolute max atx=-4.

Graphical Optimisation via Tangent Lines

• To maximise the ratio (output ÷ labour) on a plot of output vs.
labour hours starting at the origin, draw a line from the origin that is tangent to the curve.
• The tangency point delivers the greatest slope → optimal efficiency.

Practical & Pedagogical Reminders

• Always factor f'' when possible—makes sign charts and the second-derivative test quick and transparent.
• Multiplicative factors may be cancelled for sign analysis but never cancel factors joined by addition/subtraction.
• When a derivative is undefined yet the original function is defined, the corresponding x-value is a critical number.
• “Pull everything into multiplicative pieces” before conducting the second-derivative test on complicated products.
• E-functions: \frac{d}{dx}(e^{x})=e^{x}$$ ⇒ product rule often repeats the same factor.
• Chain-rule powers: keep the inside untouched, multiply by the derivative of the inside before simplifying.

End-of-Lecture Logistics

• All lecture Flex assignments posted; homework extended to Sunday (late video release).
• Exam 2 scheduled for next Friday — cover increasing/decreasing, relative & absolute extrema, concavity, higher derivatives. Study the second-derivative test and EV Theorem thoroughly.