Maximum and Minimum Values section 4.1 Definition. Consider a function f on its domain D. (i) We say that f has absolute maximum at a point x 0 D if for all x D we have f(x) f(x 0 ). (ii) We say that f has absolute minimum at a point x 0 D if for all x D we have f(x) f(x 0 ). For example, in the following figure which is taken from the textbook, the value f(a) is the absolute minimum and f(d) is the absolute maximum. Question. Under what circumstances is it guaranteed that a function has absolute extrema on its domain? The following is an answer to this question. The Extreme Value Theorem. If f is continuous on a closed interval [a,b], then it has an absolute maximum and an absolute minimum on [a, b].
Question. Now that we know the circumstances, how could we actually search for the points of absolute extrema?. The following theorem answers this question: Theorem. The f(x 0 ) is an absolute extrema of function on an interval I, then x 0 is amongst one of the following points: x 0 is an endpoint of the interval I. x 0 is a critical number of f Closed Interval Method. To find the absolute maximum and minimum values of a continuous function f on a closed interval [a, b]: Find the values of f at the points of [a, b] where either the derivative is zero or the derivative does not exist (the critical numbers). Find the values of the function at the endpoints a and b. Compare these values from these two categories to declare the maximum and minimum values. Page 2
Example (From the final exam of summer 2016). Determine the absolute maximum and minimum of f(x) = 1 3 x3 +4x on the interval [0, 3]. Also, verify that the function satisfies the hypothesis of the Extreme Value Theorem. Solution: Verification of the Extreme Value Theorem: f is continuous since it is a polynomial, and the domain is a closed interval, so the conditions of the Extreme Value Theorem are met. (this part gets some marks). Now, use the Closed Interval Method: (the discussion below gets the remaining part of the grade) Next Step: Find the derivative: f (x) = x 2 +4 Next Step: Search for the points in the domain at which the derivative does not exist: there are no such points. Next Step: Search for the points in the domain at which the derivative is zero: f (x) = x 2 +4 = 0 x 2 = 4 x = ±2 only 2 is in the domain Next Step: According to the previous two steps, the only critical number is x = 2. Next Step: Calculate the value of f at the the critical numbers and the endpoints and then compare these values: Page 3
f(2) = 1 3 23 +4(2) = 8 3 +8 = 16 3 f(0) = 0 f(3) = 1 3 33 +4(3) = 9+12 = 3 Therefore, f has the absolute maximum of 16 3 and has the absolute minimum of 0. Page 4
Example (From the final exam of Fall 2016). Determine the absolute maximum and minimum of f(x) = x 3 3 2 x2 6x over the interval [ 2, 3]. Solution: Verification of the Extreme Value Theorem: f is continuous since it is a polynomial, and the domain is a closed interval, so the conditions of the Extreme Value Theorem are met. (this part gets some marks). Now, use the Closed Interval Method: (the discussion below gets the remaining part of the grade) Next Step: Find the derivative: f (x) = 3x 2 3x 6 = 3(x 2 x 2) = 3(x+1)(x 2) Next Step: Search for the points in the domain at which the derivative does not exist: there are no such points. Next Step: Search for the points in the domain at which the derivative is zero: f (x) = 0 x = 1, 2 both are in the domain Next Step: According to the previous two steps, the only critical numbers are x = 1, 2. Next Step: Calculate the value of f at the the critical numbers and the endpoints and then compare these values: Page 5
f( 1) = 7 2 f(2) = 10 f( 2) = 2 f(3) = 9 2 absolute maximum absolute minimum Page 6
Example (section 4.1 exercise 49). Use the Closed Interval Method to find Determine the absolute maximum and minimum of f(x) = 2x 3 3x 2 12x+1 on the interval [ 2, 3]. Also, verify that the function satisfies the hypothesis of the Extreme Value Theorem. Solution: Verification of the Extreme Value Theorem: f is continuous since it is a polynomial, and the domain is a closed interval, so the conditions of the Extreme Value Theorem are met (this part gets some marks). Now, use the closed interval method: (the discussion below gets the remaining part of the grade) Next Step: Find the derivative: f (x) = 6x 2 6x 12 = 6(x 2 x 2) = 6(x+1)(x 2) Next Step: Search for the points in the domain at which the derivative does not exist: there are no such points. Next Step: Search for the points in the domain at which the derivative is zero: f (x) = 0 x = 1, 2 both are in the domain Next Step: According to the previous two steps, the only critical numbers are x = 1, 2. Next Step: Calculate the value of f at the the critical numbers and the endpoints and then compare these values: Page 7
candidate x f(x) 1 8 absolute max 2-19 absolute min 2 3 3 8 Page 8
Example (section 4.7 exercise 3). Determine the absolute maximum and minimum of. f(x) = x + 1 x over the interval 1 2 x 5. Also, verify that the function satisfies the hypothesis of the Extreme Value Theorem. Solution: Verification of the Extreme Value Theorem: f is continuous since the only point which makes the function non-continuous is the point x = 0, but this point is not in the domain of f given in this question. Secondly, the domain is a closed interval, so both conditions of the Extreme Value Theorem are met. (this part gets some marks). Now, use the closed interval method: (the discussion below gets the remaining part of the grade) Next Step: Find the derivative: f (x) = 1 x 2 = 1 1 x 2 Next Step: Search for the points in the domain at which the derivative does not exist: The only point at which this derivative does not exists is the point x = 0 but this point is not in the domain so it is ruled out. Next Step: Search for the points in the domain at which the derivative is zero: f (x) = 0 1 1 x 2 = 0 1 x 2 = 1 x 2 = 1 x = 1 x = 1 this is not acceptable ; not in the domain Next Step: According to the previous two steps, the only critical number is x = 1. Next Step: Calculate the value of f at the the critical numbers and the endpoints and then compare these values: Page 9
candidate x f(x) 1 2 absolute min 1 2 2.5 5 5.2 absolute max Page 10