Let h(x) = e + 2x.Use the Intermediate Value Theorem to show that e" + 2x :0 has at least one solution.Use a proof by contradiction together with part (a) and Rolle's Theorem to show thatet + 2x0 has exactly one solution.(Note: A different way to prove part (b) is to show that h(x) is always increasing.)

Answered: Let h(x) = e" + 2x. Use the…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.3: Zeros Of Polynomials

Problem 4E

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Question

Let h(x) = e + 2x.
Use the Intermediate Value Theorem to show that e" + 2x :
0 has at least one solution.
Use a proof by contradiction together with part (a) and Rolle's Theorem to show that
et + 2x
0 has exactly one solution.
(Note: A different way to prove part (b) is to show that h(x) is always increasing.)

Expert Solution

Step 1

First it is required to show that h(x)=0 has at least one solution, where

$h (x) = e^{x} + 2 x$

Step 2

Note that according to the Intermediate Value Theorem if a function is continuous whose domain consist the interval [a, b], then the function takes on every value between f(a) and f(b).

Here, the given function h(x) is continuous function for every real value of x (Sum of two continuous function), and domain of h(x) contain the interval [-5, 5] , and

$f (- 5) = e^{(- 5)} + 2 (- 5) = - 9.99, a n d f (5) = e^{5} + 2 (5) = 158.41 A n d 0 \in (f (- 5), f (5)) T h u s, t h e r e e x i s t a c f o r w h i c h e^{c} + 2 c = 0 T h u s, t h e r e e x i s t a t l e a s t o n e s o l u t i o n f o r h (x) = e^{x} + 2 x = 0$

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