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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 27e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 27e

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# Locating critical points a. Find the critical | Ch 4.1 - 27E

ISBN: 9780321570567 2

## Solution for problem 27E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition

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Problem 27E

Locating critical points a. Find the critical points of the following functions on the domain or on the given interval. b. Use a graphing unity In determine whether each critical point corresponds to a local minimum. local maximum. or neither. x ?x f(x) = e +e 2

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STEP_BY_STEP SOLUTION Step-1 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 1 1 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Example ; Find all critical points of f(x) = x 8x . 2 Because f (x) is a polynomial function, its domain is all real numbers. 1 f (x) = 4x 16x f (x) = 0 =4x 16x 4x ( x - 4) = 0 4x(x+2)(x-2) = 0 , since a - b = (a+b)(a-b) X = 0 , x = -2 and x=2. 4 2 At x= -2 , then f(-2) = (2) 8(2) = 16 - 32 = -16. At x= 0 , then f(0) = (0) 8(0) = 0 - 0 = 0. At x= 2 , then f(2) = (2) 8(2) = 16 - 32 = -16. Hence, the critical points of f(x) are (2,16), (0,0), and (2,16). Step-2 e +e x a). The given equation is ; f(x) = 2 , here f(x) is exponential function , its domain is all real numbers .so, f(x) is continuous for all of x. e +ex Now , f(x) = 2 then differentiate the function both sides with respect to x. x x So , f (x) = d ( e +e ) dx 2 dx(e ) dx(ex ) d x x = 2 since dx(e ) = e e e x = ( 2 ) 1 ec e c Since from the definition f (c) = 0 = ( 2 ) 1 c 2c (2) e (1e ) = 0. Now , there are two equations to solve That is , e = 0 and(1e 2c ) = 0. solve these two equations and find the value of c. If , e = 0 c Taking log on both sides , then log( e ) = log(0) c(loge) =log(0) c(1) = , since loge = 1. If , (1e 2c ) = 0.Taking log on both sides , then log(1) = log( e 2c(loge) = 0 , since loge = 1, and log(1) =0 , log( a ) = m(loga) 2c =0 c=0. Therefore , f(x) attains the critical value at x = 0. e +e0 At , x= 0 then f(0) = = 1+1 , since e = 1. 2 2 Hence , the critical point is (0,1) Step-3 x x b). The graph of the related function f(x) = e +e is; 2 Therefore, from the graph it is clear that the function is decreasing in the negative interval up to zero , from zero to positive interval it is increasing . So, the function attains the local minimum at x=0.That is the function attains the local minimum at the critical point (0,1).

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