Suppose f(x) = x + 3x + 1. In this problem, we will show that f has exactly one root (or zero) in the interval [-4,-1]. (a) First, we show that f has a root in the interval (-4,-1). Since f is a choose function on the interval [-4,-1] and f(-4) and f(-1) = the graph of y = f(x) must cross the x-axis at some point in the interval (-4,-1) by the choose ◆. Thus, f has at least one root in the interval [-4, -1]. = (b) Second, we show that f cannot have more than one root in the interval [—4, -1] by a thought experiment. Suppose that there were two roots x = a and x = b in the interval [-4, -1] with a
Suppose f(x) = x + 3x + 1. In this problem, we will show that f has exactly one root (or zero) in the interval [-4,-1]. (a) First, we show that f has a root in the interval (-4,-1). Since f is a choose function on the interval [-4,-1] and f(-4) and f(-1) = the graph of y = f(x) must cross the x-axis at some point in the interval (-4,-1) by the choose ◆. Thus, f has at least one root in the interval [-4, -1]. = (b) Second, we show that f cannot have more than one root in the interval [—4, -1] by a thought experiment. Suppose that there were two roots x = a and x = b in the interval [-4, -1] with a
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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![Suppose f(x) = x² + 3x + 1. In this problem, we
will show that f has exactly one root (or zero) in the
interval [-4,-1].
(a) First, we show that f has a root in the interval
(-4,-1). Since f is a choose
function on
the interval [-4, -1] and f(-4)=
and
f(-1) =, the graph of y = f(x) must cross the
x-axis at some point in the interval (-4,-1) by the
◆. Thus, f has at least
choose
one root in the interval [—4, -1].
(b) Second, we show that f cannot have more than
one root in the interval [−4, -1] by a thought
experiment. Suppose that there were two roots
x = a and x = b in the interval [-4, −1] with a < b.
Then f(a) = f(b) = Since fis
choose
on the interval [-4,-1] and
choose
on the interval (-4,-1), by
choose
there would exist a
point c in interval (a, b) so that f'(c) = 0. However,
the only solution to f'(x) = 0 is x =
not in the interval (a, b), or in [−4, -1]. Thus, f
cannot have more than one root in [−4, −1].
0
which is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc3231216-ac6a-463c-8e67-724e263e6aaa%2F39abdae6-5afc-4719-9d05-0c732c220c05%2F0y21icq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Suppose f(x) = x² + 3x + 1. In this problem, we
will show that f has exactly one root (or zero) in the
interval [-4,-1].
(a) First, we show that f has a root in the interval
(-4,-1). Since f is a choose
function on
the interval [-4, -1] and f(-4)=
and
f(-1) =, the graph of y = f(x) must cross the
x-axis at some point in the interval (-4,-1) by the
◆. Thus, f has at least
choose
one root in the interval [—4, -1].
(b) Second, we show that f cannot have more than
one root in the interval [−4, -1] by a thought
experiment. Suppose that there were two roots
x = a and x = b in the interval [-4, −1] with a < b.
Then f(a) = f(b) = Since fis
choose
on the interval [-4,-1] and
choose
on the interval (-4,-1), by
choose
there would exist a
point c in interval (a, b) so that f'(c) = 0. However,
the only solution to f'(x) = 0 is x =
not in the interval (a, b), or in [−4, -1]. Thus, f
cannot have more than one root in [−4, −1].
0
which is
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