×
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 11e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 11e

×

# maximum/minimum values from graphs Use the following

ISBN: 9780321570567 2

## Solution for problem 11E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants

Calculus: Early Transcendentals | 1st Edition

4 5 1 336 Reviews
19
5
Problem 11E

maximum/minimum values from graphs ?Use the? ?following graphs to identify the points? (?if any?) ?on the interval? [?a. b?] ?at which the function has cm absolute maximum value or an absolute minimum value.

Step-by-Step Solution:
Step 1 of 3

STEP_BY_STEP SOLUTION Step-1 When an output value of a function is a maximum or a minimum over the entire domain of the function, the value is called the absolute maximum or the absolute minimum. Let f be a function with domain D and let c be a fixed c onstant in D. Then the output value f ) is the 1. Absolute maximum value of f on D if and only if f(x) f(c) , for all x in D. 2. Absolute minimum value of f on D if and only if f(c) f(x) , for all x in D. Example ; The Absolute extreme values on a restricted domain ; If th e domain of f(x ) = x is restricted to [-2, 3], the corresponding range is [0, 9]. As shown below, the graph on the interval[-2, 3] suggests that f has an absolute maximum of 9 at x = 3 and an absolute minimum of 0 at x = 0. Step-3 The given graph is; Now we need to verify the points , from the graph on the interval [a,b] at which the function has an absolute maximum value or absolute minimum value. From the graph it is clear that the given function y = h(x) is continuous on that given interval [a,b]. Since the graph has no holes or breaks in that interval. So, the function is continuous on [a,b]. We know the result that a function is continuous on the closed interval [a,b] has an absolute maximum value and an absolute minimum value on that interval . Hence , the graph y = h(x) is continuous on [a, b] . So , by the above results maximum value occurs at x= b , and the minimum value occurs at x= c 2 on [a, b].Since it is clearly mentioned in the graph. NOTE : Let f be a continuous function defined on an open interval containing a number ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) = 0 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). To determine the absolute maximum and minimum values of a continuous function on a closed interval; That is ; 1. Verify that the function is continuous on the interval [a, b]. 2. Find all critical points of f(x) that are in the interval (a,b). 3. Add the endpoints a and b of the interval [a,b] to the list of points found in step-2. 4. Compute the value of f at each of the points in this list. 5. The largest value in step 4 is the absolute maximum value of f on [a, b] and the smallest value is the absolute minimum value. Example;Determine the absolute extreme values for the following function and interval. g(t) = 2t +3t -12t +4 0n[-4,2]. 1 2 g (t) = 0 = 6t +6t-12 2 6( t +t -2) = 0. 2 ( t +2t t-2) = 0. t ( t+2) - (t+2) = 0. (t+2)(t-1) = 0 t= -2 , and t=1. Therefore , critical values are t =-2, 1, and the end points are -4,2. At , t= -4 then g(-4) = 2(4) +3(4) -12(-4) +4 = -128+48+48+4 = -28. At , t= -2 then g(-2) = 2(2) +3(2) -12(-2) +4 = -16+12+24+4 =24. At , t= 1 then g(1) = 2(1) +3(1) -12(1) +4 = 2+3-12+4 = -3. At , t= 2 then g(2) = 2(2) +3(2) -12(2) +4 = 16+12-24+4 = 8. Hence , the absolute maximum of g(t) is 24 and it occurs at t = -2 ( a critical point), and the absolute minimum of g(t) is -28 and it occurs at t = -4 (an endpoint).

Step 2 of 3

Step 3 of 3

#### Related chapters

Unlock Textbook Solution