Download original notes
PAGE 1

4.1 Maximum and Minimum Values

Goal: Find max/min values of a continuous function over some interval \( a \leq x \leq b \).

Possible Situations

In the first scenario, we identify the absolute extrema on the interval \( [a, b] \). The highest point on the curve corresponds to the absolute (or global) maximum, while the lowest point corresponds to the absolute (or global) minimum.

Graph of a continuous function on [a, b] showing absolute maximum at the right endpoint and absolute minimum at a local valley.

In the second scenario, we distinguish between local and absolute extrema. A peak within the interval is a relative or local maximum, while the highest overall point is the absolute maximum. Similarly, a valley is a relative or local minimum, and the lowest overall point is the absolute minimum.

Graph showing local maximum, local minimum, absolute maximum at an endpoint, and absolute minimum at a valley.

Extrema can also occur at sharp corners or endpoints. In these examples, the absolute maximum and minimum are clearly marked at specific points on the piecewise or linear segments within the interval \( [a, b] \).

V-shaped graph on [a, b] with absolute maximum at an endpoint and absolute minimum at the vertex.
Graph with a sharp peak showing absolute maximum at the peak and absolute minimum at the right endpoint.
PAGE 2

Possible Locations of Absolute Extrema

We see that the possible locations of absolute max/min are:

  • End points of the interval
  • Where tangent line is horizontal \( (f' = 0) \)
  • Where the slope of tangent line is not defined \( (f' \text{ DNE}) \)

Procedure

So, given \( f(x) \) and interval \( a \leq x \leq b \):

Find the above locations then compare values of \( f(x) \).

PAGE 3

Example: Absolute Maximum and Minimum Values

Find the abs. max/min values of

\[ f(x) = 2x^3 - 6x^2 - 18x + 4 \quad \text{on} \quad -2 \le x \le 4 \]

Find values of \( x \) where \( f' = 0 \) or \( f' \text{ DNE} \).

\[ f'(x) = 6x^2 - 12x - 18 \]\[ f' = 0 \rightarrow 6x^2 - 12x - 18 = 0 \]\[ x^2 - 2x - 3 = 0 \]\[ (x - 3)(x + 1) = 0 \]

\( x = -1, \quad x = 3 \)

Critical numbers (where \( f' = 0 \) or \( f' \text{ DNE} \)) possible locations of max/min

\( f' \text{ DNE} \) does not happen since \( f(x) \) and \( f'(x) \) are polynomials.

Possible locations of max/min values:

  • \( x = -2 \), \( x = 4 \) — end points of interval
  • \( x = -1 \), \( x = 3 \) — critical numbers

Now we compare \( f(x) \) at these locations.

PAGE 4
\[ f(-2) = 2(-2)^3 - 6(-2)^2 - 18(-2) + 4 = 0 \]\[ f(-1) = 14 \]\[ f(3) = -50 \]\[ f(4) = -36 \]

\( f(-1) = 14 \rightarrow \) largest, so this means abs. max value of \( f(x) \) is 14 at \( x = -1 \)

\( f(3) = -50 \rightarrow \) smallest, so abs. min \( = -50 \) at \( x = 3 \)

PAGE 5

Example: Finding Absolute Extrema

Consider the function:

\[ f(x) = x + \frac{1}{x} \text{ on } 1 \le x \le 4 \]

Interval we care about: \( 1 \le x \le 4 \)

Rewrite the function for easier differentiation:

\[ f(x) = x + x^{-1} \]

Find Critical Numbers

Critical numbers occur where \( f'(x) = 0 \) or \( f'(x) \) is DNE (Does Not Exist).

\[ f'(x) = 1 - x^{-2} = 1 - \frac{1}{x^2} \]

Case 1: \( f'(x) = 0 \)

\[ 1 - \frac{1}{x^2} = 0 \implies 1 = \frac{1}{x^2} \implies x^2 = 1 \]

This gives two potential values: \( x = 1 \) and \( x = -1 \).

  • Keep \( x = 1 \)
  • Discard \( x = -1 \): Falls outside the interval.

Case 2: \( f'(x) \) DNE

\[ 1 - \frac{1}{x^2} = f' \]

\( f' \) is DNE at \( x = 0 \).

Discard \( x = 0 \) because:

  1. Outside interval
  2. \( f(x) \) is not defined at \( x = 0 \)

Compare Values at Critical Points and Endpoints

So we keep only one critical number: \( x = 1 \) (which happens to be the left endpoint). Now compare \( f(x) \) at the endpoints:

\[ f(1) = 1 + \frac{1}{1} = 2 \quad \text{Absolute minimum at } x = 1 \]\[ f(4) = 4 + \frac{1}{4} = \frac{17}{4} \quad \text{Absolute maximum at } x = 4 \]
PAGE 6

Example: Finding Critical Numbers with Inverse Trig

Consider the function:

\[ f(x) = 2x^2 + 2\cos^{-1}x \text{ on } -1 \le x \le 1 \]

Find Critical Numbers

Find where \( f'(x) = 0 \) or \( f'(x) \) is DNE.

\[ f'(x) = 4x + 2 \cdot \frac{-1}{\sqrt{1-x^2}} = 4x - \frac{2}{\sqrt{1-x^2}} = \frac{4x\sqrt{1-x^2} - 2}{\sqrt{1-x^2}} \]

Solving \( f'(x) = 0 \)

A fraction equals zero when its numerator equals zero:

\[ 4x\sqrt{1-x^2} - 2 = 0 \]\[ 4x\sqrt{1-x^2} = 2 \]\[ (4x\sqrt{1-x^2})^2 = 2^2 \]\[ 16x^2(1-x^2) = 4 \]\[ 16x^2 - 16x^4 - 4 = 0 \]

Divide by \( -4 \):

\[ 4x^4 - 4x^2 + 1 = 0 \]

This is a quadratic equation in terms of \( x^2 \):

\[ 4(x^2)^2 - 4(x^2) + 1 = 0 \]\[ (2x^2 - 1)(2x^2 - 1) = 0 \]
PAGE 7
\[ 2x^2 - 1 = 0 \]\[ x^2 = \frac{1}{2} \]
\( x = \frac{1}{\sqrt{2}} \)
\( x = -\frac{1}{\sqrt{2}} \)

are they both inside the interval we care about? yes, so keep them

Finding where \( f' \) Does Not Exist (DNE)

\( f' \text{ DNE} \rightarrow \text{fraction DNE} \rightarrow \text{denominator} = 0 \)

\[ \sqrt{1-x^2} = 0 \]\[ 1-x^2 = 0 \]\[ x=1, \quad x=-1 \]

they are the end points of interval, so keep them

Comparing Function Values

now compare \( f(x) \) (NOT \( f'(x) \)) at \( x = -1, \) \( x = -\frac{1}{\sqrt{2}}, \) \( x = \frac{1}{\sqrt{2}}, \) \( x = 1 \)

\( f(-1) = 2(-1)^2 + 2 \cos^{-1}(-1) = 2 + 2(\pi) = 2 + 2\pi \)

abs. max at \( x = -1 \)

\( f(-\frac{1}{\sqrt{2}}) = 2(-\frac{1}{\sqrt{2}})^2 + 2 \cos^{-1}(-\frac{1}{\sqrt{2}}) = 1 + 2(\frac{3\pi}{4}) = 1 + \frac{3\pi}{2} \)

\( f(\frac{1}{\sqrt{2}}) = 2(\frac{1}{\sqrt{2}})^2 + 2 \cos^{-1}(\frac{1}{\sqrt{2}}) = 1 + 2(\frac{\pi}{4}) = 1 + \frac{\pi}{2} \)

\( f(1) = 2(1)^2 + 2 \cos^{-1}(1) = 2 + 2(0) = 2 \)

abs. min at \( x = 1 \)