1. Ch. 5 – Applications of Derivatives 5.1 – Extreme Values of Functions

2. Absolute Extreme Values: Let f be a function with domain D. Then f(c) is the... –...absolute maximum value on D if f(x) ≤ f(c) for all x in D. –...absolute minimum value on D if f(x) ≥ f(c) for all x in D. – The absolute max and min values can be endpoints or interior points of a domain. Ex: Use your grapher to find the x-coordinates of the absolute extrema for the following functions: – Absolute min at x = -2 – Absolute max at x = 1 – Absolute min at x = π – No absolute max because x = 0 and x = 2π aren’t in the domain

3. Extreme Value Theorem: If f is continuous on a closed interval [a, b], then f has both a maximum and a minimum on the interval. Local (or Relative) Extrema: – A local maximum occurs at x = c when f(c) is larger than the f(x) values to its left and right. – A local minimum occurs at x = c when f(c) is smaller than the f(x) values to its left and right. To find the local extrema algebraically, find all points c such that f’(c) = 0 or f’ does not exist and find the values of the endpoints.

4. Ex: Use a grapher to find the extrema over the interval [-1, 2] for f(x) = x 4 – x 3 – x 2 + x + 1. – Use zstandard, then change the domain to [-5, 5] – We have several extrema here. Use your calc menu to find any local maxima and minima in the interval [-1, 2] Endpoints: (-1, 1) and (2, 7)  THESE ARE LOCAL EXTREMA! Local maxima: (.390, 1.201), (-1, 1), and (2, 7) Local minima: (-.640,.380) and (1, 1) – Absolute minimum: (-.640,.380)  LOWEST Y- VALUE! – Absolute maximum: (2, 7)  HIGHEST Y-VALUE! – If we graph f’, we see that the local extrema (excluding endpoints) of f are the same as the zeros of f’

5. A critical point is a point in the interior of the domain of a function f at which f’=0 or f’ does not exist. – A stationary point is a critical point where f’=0. Ex: Find the absolute extrema of f(x)=3x 2 – 12x + 5 by hand over the interval [0, 3]. – First, find the endpoints: Endpoints: (0, 5) and (3, -4) – Second, find the critical points: f‘ = 6x – 12 = 0 at x = 2......and f’(x) is defined for all x over [0, 3], so the only critical value is at (2, -7) – With all of these points, find your absolute extrema! Absolute min: (2, -7) Absolute max: (0, 5)

6. Use a slope sign chart to find the x-coordinates of the local extrema for f(x) = x 3 - 4x 2 - 3x +2. – No endpoints here, so check critical points... – Now make a number line with the 2 critical points. This number line represents the domain of f’. – We’ve divided the domain of x into 3 sections; choose an x-value from each section and plug them into f’ to determine whether or not the slope of f is positive or negative. If the value is positive (or negative), mark a plus (or a minus) in that section of the number line – Since the sign chart represents the slopes of f......we know we have a local maximum at x = -1/3 because the slope goes from positive to negative...we know we have a local minimum at x = 3 because the slope goes from negative to positive ++ -