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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.3 - Problem 30e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.3 - Problem 30e

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# More graphing Make a complete graph of the | Ch 4.3 - 30E ISBN: 9780321570567 2

## Solution for problem 30E Chapter 4.3

Calculus: Early Transcendentals | 1st Edition

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Problem 30E

More graphing ?Make a complete graph of the following functions. If an interval is not specified, graph the function on its domain. Use a graphing utility to check your work.

Step-by-Step Solution:
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Solution:- Step1 Given function is x2 2 g(x)= e And the domain of given function is R x2 Because x takes all real values and e 2 is always positive. In given function x powers are an even . So, the given function is symmetrical about y-axis. Step2 Differentiate the given equation to find g’(x) we get, 2 x g’(x)= e 2 * *(-2x) 2 x2 = e 2 (-x) x2 2 =(-x)e Again differentiate g’(x) to find g”(x) we get, x x2 ( x)e 2.x + e 2 g”(x)= - { } 2 2 x x = e 2.x e 2 x2 2 2 = (x -1)e Step3 To get extreme value we have to use g’(x)=0 x2 2 ( x)e =0 x 2 x=0 since e =/ 0 x2 x -1=0 since e 2 =/ 0 anywhere x =1 x=±1 Therefore inflection points are -1 and 1 Step4 We have to find increasing, decreasing and concavity x=0 is the critical point. Therefore in between (-,0) the function g(x) is increasing and ((,0) it is decreasing. x Let g”(x)=(x -1)e 2 Inflection points are x=-1, x=1 Therefore function g(x) is concave downwards on(-1,1). Step 5 Extreme values and inflection points. x=0 is the critical point. x2 g”(x)=(x -1)e 2 At x=0 g”(x)<0 Therefore at x=0 g(x) has maximum value . Step 6 Asymptotes and end behavior Let g’(x)=0 x=0 At x=0, g(x) then tangent parallel to x- axis x2 2 limg(x)=lim e = 1 x1 x1 x lim g(x)= lim e 2 = 0 x x Intercepts At x=0 y=1 If y=0 2 x e 2 =0 is not possible Therefore the curve do not touched y axis. Step 7 Graph of given equation

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##### ISBN: 9780321570567

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