RC

4.1.2 Maxima and Minima

Visualizing Local Maxima & Minima

  • Begin with the geometric intuition:
    • Draw shrinking secant lines around a peak or trough.
    • As the two points approach one another the secant becomes the tangent.
    • At a true peak/trough (local extremum) that tangent line is horizontal ⇒ slope = 0.
  • Immediate consequence: Wherever a differentiable function has a local max or min, its derivative must vanish.
  • Vocabulary reminder:
    • Local (relative) maximum – height greater than all nearby points.
    • Local (relative) minimum – height smaller than all nearby points.
    • Absolute (global) maximum/minimum – highest/lowest on the entire domain or on a stated interval.

Local Extreme Value Theorem

  • Formal statement:
    • If f has a local extremum at x=c and f'(c) exists, then f'(c)=0.
  • Significance:
    • Gives a necessary condition but not a sufficient one.
    • Explains why horizontal tangents are systematically sought when hunting for extrema.

Definition: Critical Point (a.k.a. Critical Number)

  • Conditions (must be an interior point of the domain):
    1. f'(c)=0 or
    2. f'(c) fails to exist (DNE).
  • Critical points are the universal “candidates” for local extrema. They tell us where to look but not what we will find.

Two canonical shapes where f' fails and an extremum still occurs

  • Sharp cusp/vertex that comes to a point → derivative undefined, yet a local max/min is present.
  • Vertical tangent that changes from increasing to decreasing.

Two shapes showing that not every critical point is an extremum

  • Flat inflection: f'(c)=0 but the graph keeps increasing/decreasing through the flat spot.
  • Corner/undefined derivative where the monotone direction doesn’t switch.

Example 1 – Critical Points of f(x)=x^2\ln x (domain x>0)

  • Differentiate using the product rule.
    \begin{aligned}
    f'(x) &= 2x\ln x + x^2\cdot\frac{1}{x} \
    &= 2x\ln x + x \
    &= x\bigl(2\ln x + 1\bigr)
    \end{aligned}
  • Set f'(x)=0 → x\bigl(2\ln x + 1\bigr)=0.
    • x=0 is not in the domain, so discard.
    • Solve 2\ln x = -1 \;\Rightarrow\; \ln x = -\tfrac12 \;\Rightarrow\; x=e^{-1/2}=\dfrac{1}{\sqrt e}.
  • Only critical point: x=\dfrac{1}{\sqrt e}\approx0.6065.
  • Graph inspection (or 2nd derivative test) confirms this point is both the local and absolute minimum.

Closed-Interval Absolute Extrema Procedure (Official Recipe)

  1. Verify f is continuous on [a,b] (Extreme Value Theorem pre-condition).
  2. Find all critical points in (a,b): solve f'(x)=0 and note where f' DNE.
  3. Evaluate f at every critical point and at both endpoints a,b.
  4. Largest value among those is the absolute maximum; smallest is the absolute minimum.

Example 2 (A): f(x)=x^4-2x^3 on [-2,2]

  • Step 1 (continuous polynomial ✔️).
  • Step 2 – derivative & critical numbers:
    f'(x)=4x^3-6x^2=2x^2(2x-3) →
    f'(x)=0\Rightarrow x=0 \text{ or } x=\tfrac32.
  • Step 3 – evaluate f:
    • f(-2)=(-2)^4-2(-2)^3=16+16=32.
    • f(0)=0.
    • f(3/2)=(\tfrac32)^4-2(\tfrac32)^3!=\tfrac{81}{16}-2\cdot\tfrac{27}{8}=\tfrac{81}{16}-\tfrac{27}{4}=\tfrac{81}{16}-\tfrac{108}{16}=-\tfrac{27}{16}\approx-1.6875.
    • f(2)=2^4-2(2)^3=16-16=0.
  • Step 4 – compare values:
    • Absolute max =32 at x=-2 (left endpoint).
    • Absolute min =-\dfrac{27}{16} at x=\tfrac32 (critical point).
    • Note: x=0 (critical) gives horizontal tangent but is neither a local nor global extremum.

Graph commentary: peak at $(-2,32)$, valley at $(1.5,-1.6875)$, gentle flattening at $(0,0)$ before rising again.

Example 3 (B): g(x)=x^{2/3}(2-x) on [0,2]

  • Preliminary rewrite to single power terms: g(x)=2x^{2/3}-x^{5/3}.
  • Differentiate:
    \begin{aligned}
    g'(x) &= \frac43 x^{-1/3}-\frac53 x^{2/3}\
    &= x^{-1/3}\Bigl(\tfrac43-\tfrac53 x\Bigr).
    \end{aligned}
  • Critical numbers inside (0,2):
    1. Factor x^{-1/3}=0 impossible, but x=0 is interior endpoint? It’s actually an ENDPOINT; derivative fails to exist there, so x=0 enters as a candidate.
    2. Solve \tfrac43-\tfrac53 x=0 ⇒ x=\dfrac{4}{5}=0.8.
  • Evaluate g at candidates and endpoints x=0,0.8,2:
    • g(0)=0 (corner, derivative DNE).
    • g(0.8)= (0.8)^{2/3}(2-0.8)\approx 1.095 – a local maximum.
    • g(2)=0.
  • Largest function value: \approx1.095 at x=4/5 ⇒ absolute max.
  • Smallest value: 0 occurs twice (both endpoints) ⇒ absolute minima at x=0 & x=2.
  • Graph check: symmetric‐looking hump with flat zeros at both ends and a pointed local max at x=0.8.

Conceptual & Practical Implications

  • Differentiability is not required for extrema; only continuity is required for absolute extrema on a closed interval.
  • Critical numbers streamline optimization: you never have to check any other interior points.
  • Real-world translations:
    • Economics → profit maximization: set marginal profit =0 but also verify corners/constraints.
    • Physics → equilibrium positions where net force (derivative of potential) is zero or undefined (e.g.
      piecewise potentials).
  • Ethical / modeling caution: a critical number that’s neither max nor min often signals an inflection or plateau; decisions based solely on f'(x)=0 can mislead if the second derivative or direct evaluation is skipped.

Summary Checklist

  • To locate local extrema: find all critical points, then apply first or second derivative tests.
  • To locate absolute extrema on a closed interval:
    1. Confirm continuity.
    2. Compute all critical points.
    3. Evaluate f at critical points and endpoints.
    4. Compare values – largest ⇒ max, smallest ⇒ min.
  • Remember: f'(c)=0 is necessary (when f' exists) but not sufficient for an extremum.
  • Non-differentiable points (cusps, vertical tangents) are equally critical and deserve the same scrutiny.