Maximum and Minimum Values section 4.1

Transcription

1 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 I if for all x I we have f(x) f(x 0 ). 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 Extrem 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 an interior point at which the derivative does not exist. x 0 is an interior point at which we have f (x 0 ) 0. Note. In the above theorem, the points which fall into the second and third categories are called critical points. Geometrically, these are the points at which either a tangent line does not exist or the tangent line is horizontal. 1

2 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. Example. Find the absolute extrema of the function f(x) x + 1 x over the interval 1 x 5 if they exist. Step 1. Because the function is continuous over the closed interval [ 1, 5], according the above theorem the absolute extrema exist. Now we find them: Step. f(x) x + x 1 f (x) 1 x 1 1 x Step 3. f (x) x 1 x x x 1 x 1 this is not acceptable ; not in the domain According to the theorem, one of the candidates is x 1 and two others are the endpoints x 1 and x 5. Other candidates are the points of the interval at which the derivative does not exist, but the function is non-differentiable at the origin only and the origin is not in the domain. So the only candidates are x 1 and x 1 Step 4. Now the table: and x 5.

3 candidate x f(x) absolute min 5 5. absolute max Example. Find the absolute extrema of the function f(x) x x + 1 over the interval 1 x 1 if they exist. Step 1. Because the function is continuous over the closed interval [ 1, 1], according the above theorem the absolute extrema exist. Now we find them: Step. f(x) x(x + 1) 1 f (x) { } { } { }{ } x (x + 1) 1 + x (x + 1) 1 { }{ } { } 1 (x + 1) 1 + x}{ 1 (x + 1) 1 x x x+1 ( x+1) +x x+1 (x+1)+x x+1 3x+ x+1 Step 3. f (x) 0 3x + 0 x 3 3

4 According to the theorem, one of the candidates is x 3 and two others are the endpoints x 1 and x 1. Other candidates are the points of the interval at which the derivative does not exist, but the derivative f (x) 3x+ x+1 does not exist at the point x 1 only, which is already one of our candidates, so we do not get any new candidates. So finally the only candidates are x 3 Step 4. Now the table: and x 1 and x 1. candidate x f(x) absolute min absolute max Example (section 4.1 exercise 49). Find the absolute maximum and absolute minimum values of f(x) x 3 3x 1x + 1 on the interval [, 3] f (x) 6x 6x 1 6(x x ) f (x) 0 x x 0 x b ± b 4ac a 1 ± ± 3 1 Two other candidates are the endpoints: candidate x f(x) absolute max -19 absolute min 3 8 4

5 Example (section 4.1 exercise 5). Find the absolute maximum and absolute minimum values of f(x) (x 1) 3 on the interval [ 1, ] f (x) 3(x 1) (x) 6x(x 1) f (x) 0 x 0, x 1 0 x 0, x ±1 and all three are in the domain. Two other candidates are the endpoints, but we have already have -1 in the list of candidates: candidate x f(x) absolute min absolute max Example (section 4.1 exercise 56). Find the absolute maximum and absolute minimum values of f(t) 3 t(8 t) on the interval [0, 8] f(t) t 1 3 (8 t) 5

6 f (t) { } t (8 t) + t 3 (8 t) 1 3 t 3 (8 t) t t 3 3 t 3 t (8 t) 3t 3 3 t 8 4t 3 3 t f (t) 0 8 4t 0 t and this point is in the domain. The non-differentiable point is t 0 and this point is in the domain, so it is one of the candidates. The other candidates are the endpoints, t 0 and t 8. Put these point together in the table: candidate t f(t) 0 0 absolute min 6 3 absolute max 8 0 absolute min Example. Let us take the same function as in the previous question but change the interval to [ 1, 8] f(t) t 1 3 (8 t) f (t) { } t (8 t) + t 3 (8 t) 1 3 t 1 8 t 3 (8 t) t t 3 (8 t) 3t t t t 3 3 t f (t) 0 8 4t 0 t and this point is in the domain. 6

7 Two other candidates are the endpoints. Here the point t 0 is a candidate too, because we must consider the points in the domain at which the function is not differentiable. candidate x f(x) -1-9 absolute min absolute max 8 0 Definition. Consider a function f on its domain D. A critical number for f is any point c in the domain such that either f (c) 0 or f (c) does not exists. At the endpoints, one-sided derivative is meant. So, an endpoint is eligible to be a critical point if either the one-sided derivative is zero or the one-sided derivative does not exist. Example. Find the critical numbers of the function g(t) 5t 3 + t 5 3. g (t) 10 3 t t t t t 3 3 t g (0) does not exist therefore t 0 is a critical point. g (t) 0 t is also a critical point. So, we have only two critical points t 0, Example (section 4.1 exercise 39). Find the critical numbers of the function F (x) x 4 5 (x 4). 7

8 { } F (x) x 4 5 (x 4) { + x 4 5 (x 4) } 4 { } 5 x 1 5 (x 4) + x 4 5 (x 4) 4(x 4) 5 5 x + 5 x 4 (x 4) 4(x 4) + 10x(x 4) 5 5 x { } (x 4) + 5x { } 7x 8 (x 4) 5 5 (x 4) x 5 5 x Then: F (x) 0 x 4, 8 7 F (x) does not exist for x 0. So, we have only three critical points x 0, 4, 8 7 Example (section 4.1 exercise 41). Find the critical points of the function f(θ) cos θ + sin θ. f (θ) (cos θ) + (sin θ) sin θ + ( sin θ cos θ) sin θ( 1 + cos θ) sin θ 0 θ nπ f (θ) 0 or cos θ 1 or θ nπ But the case θ nπ includes the case θ nπ, therefore the final answer is: θ nπ n 0, ±1, ±,... Example. Find the critical points of the function f(x) x x 3 1 8

9 f (x) (x ) (x 3 1) (x )(x 3 1) (x 3 1) x(x3 + ) (x 3 1) Now we need to look for the points at which the derivative is zero or the derivative does not exist. f (x) 0 x 0, 3 there is no point in the domain where f is not differentiable (note that the point x 1 is not in the domain at all, so it is not considered as a critical point). So, the only critical points are x 0, 3. 9