Show that the function Lan f(x) = x³ +e³x has exactly one real root. First of all, by the Intermediate Value Theorem, f(x) has a solution in the interval (a, b) = choose an interval of any length) (you may Now suppose that f(x) has more than one real root. Then by the Mean Value Theorem, between any pair of real roots there must be some point c at which f'(x) is 0 However, we find f'(x) = and notice that f'(x) is always CA. negative. OB. positive. We conclude that f(x) has exactly one real root.
Show that the function Lan f(x) = x³ +e³x has exactly one real root. First of all, by the Intermediate Value Theorem, f(x) has a solution in the interval (a, b) = choose an interval of any length) (you may Now suppose that f(x) has more than one real root. Then by the Mean Value Theorem, between any pair of real roots there must be some point c at which f'(x) is 0 However, we find f'(x) = and notice that f'(x) is always CA. negative. OB. positive. We conclude that f(x) has exactly one real root.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Show that the function
f(x) = x³ +e³x
has exactly one real root.
First of all, by the Intermediate Value Theorem, f(x) has a solution in the interval (a, b) =
(you may
choose an interval of any length)
Now suppose that f(x) has more than one real root. Then by the Mean Value Theorem, between any pair of real roots there
must be some point c at which f'(x) is o
However, we find
f'(x) =
and notice that f'(x) is always
CA. negative.
OB. positive.
We conclude that f(x) has exactly one real root.
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