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4.1 Maximum and Minimum Values a Function F Has an Absolute Maximum at C If F(C) ≥ F(X) for All X in the Domain of F

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4.1 Maximum and Minimum Values a Function F Has an Absolute Maximum at C If F(C) ≥ F(X) for All X in the Domain of F 4.1 Maximum and Minimum Values A function f has an absolute maximum at c if f(c) ≥ f(x) for all x in the domain of f. The function value f(c) is the maximum value. A function f has an absolute minimum at c if f(c) ≤ f(x) for all x in the domain of f. The function value f(c) is the minimum value. The absolute maximum and minimum values are also called the extreme values of f. A function has a local maximum at x = a if f(a) ≥ f(x) for all x \near" a. A function has a local minimum at x = a if f(a) ≤ f(x) for all x \near" a. For a function to have a local max/min at x = a, it must be defined on BOTH sides of a. This is not true for an absolute max/min. Absolute max/mins can occur at an endpoint of an interval. Example: Find the absolute and local extrema for the following graphs. ( (x + 1)2 if − 3 < x ≤ 0 f(x) = 2 x2 − 1 if 0 < x ≤ 2 We can see from the above that if a function has an absolute max/min, it will occur at either a local max/min or at an endpoint of a given interval. 1 Very Important Fact: ALL local extrema of a function f will occur at values of x where the derivative is 0 or the derivative is undefined! The values of x in the domain of f for which f 0(x) = 0 or f 0(x) is undefined are called the critical numbers or critical values of f. However, even though all local extrema occur at critical values, this DOES NOT MEAN that every critical value produces a local max or min!! p Example: Find the critical numbers of the function f(x) = 3 x2 − x. Example: Find the critical numbers of the function f(x) = j9 − x2j. Example: Find the critical numbers of the function f(x) = x2=5(x + 2)3. 2 Extreme Value Theorem: If a function f is continuous on a closed interval [a; b], then f attains both an absolute maximum value and an absolute minimum value somewhere in [a; b]. The candidates for the location of absolute extrema are the critical values of the function and the endpoints of the interval. The Closed Interval Method for finding absolute extrema: First find all critical values, then evaluate the function at the critical values that lie in the interval and at the endpoints of the interval. The max/min of these values IS the absolute max/min. Example: Find the absolute maximum and minimum values of f(x) = x3 − 12x + 1 on the interval [−3; 5]. ex2 Example: Find the absolute maximum and minimum values of g(x) = on the interval [0; 3]. 2x + 1 3 Example: Find the absolute maximum and minimum values of g(x) = x + 2 sin x on the interval [0; π]. p Example: Find the absolute maximum and minimum values of g(x) = x2 9 − x2 on the interval [1; 3]. 4.2 The Mean Value Theorem Mean Value Theorem: If f is a continuous and differentiable function on the interval [a; b], then there exists a number c in the inteval (a; b) such that f(b) − f(a) f 0(c) = b − a Meaning: There is at least one value c where the tangent line at c is parallel to the secant line between (a; f(a)) and (b; f(b)). Show that the following functions satisfy the Mean Value Theorem on the given interval by finding the number(s) c that satisfy the conclusion of the theorem. f(x) = ln x on the interval [1; e] f(x) = x3 − 1 on the interval [−2; 2] 4.
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