Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
I would like to understand how the Taylor formula was applied how did we get those numbers and why is x in absolute value
![is
dom f = {x ER | x² + 3x = x(x+3) ≥ 0} = (-∞, -3] U [0, ∞).
Since the domain of f isn't symmetric with respect to 0, the function is neither odd nor even.
In order to compute the required limits, observe that, as x→ ±∞,
I
3
3
1 3
f(x)=√x² + 3x - |x| = |x|| ( √ ₁ + ² – 1) — 1¹ ( ¹ + 2 + ² + 0 ( - ) − ¹) -+
1+
= x 1
2
X
x
X
Therefore
3
x+∞ 2
lim f(x)= lim
x++∞
lim f(x)= lim
8118
Mac
X
7 81←H
X
3 x
x
312
lim
x++∞ 2
312
lim
x-x 2
3
NIC
2
Nic
2
=
+0(1).
The function h(x)=√x² + 3x is the composition of the polynomial P(x) = x² + 3x (which is
continuous in R) with the function g(t) = √t (continuous if t > 0), therefore it's continuous
on its domain. Since f is the sum of h with the continuous function -x, we have that f is
continuous on its domain. Therefore lim f(x) = f(-3) = 0 and lim f(x) = f(0) = 0. The
function f has no vertical asymptotes.
x-3
x-0
f has right horizontal asymptote y = 3/2 and left horizontal asymptote y = -3/2.
(b) Study the continuity and differentiability of the function, locate and classify possible points of
discontinuity and non-differentiability in dom f and compute its derivative.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F130bcfaa-134b-4437-b161-7ed0ee7cf6af%2Fk0u6gk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:is
dom f = {x ER | x² + 3x = x(x+3) ≥ 0} = (-∞, -3] U [0, ∞).
Since the domain of f isn't symmetric with respect to 0, the function is neither odd nor even.
In order to compute the required limits, observe that, as x→ ±∞,
I
3
3
1 3
f(x)=√x² + 3x - |x| = |x|| ( √ ₁ + ² – 1) — 1¹ ( ¹ + 2 + ² + 0 ( - ) − ¹) -+
1+
= x 1
2
X
x
X
Therefore
3
x+∞ 2
lim f(x)= lim
x++∞
lim f(x)= lim
8118
Mac
X
7 81←H
X
3 x
x
312
lim
x++∞ 2
312
lim
x-x 2
3
NIC
2
Nic
2
=
+0(1).
The function h(x)=√x² + 3x is the composition of the polynomial P(x) = x² + 3x (which is
continuous in R) with the function g(t) = √t (continuous if t > 0), therefore it's continuous
on its domain. Since f is the sum of h with the continuous function -x, we have that f is
continuous on its domain. Therefore lim f(x) = f(-3) = 0 and lim f(x) = f(0) = 0. The
function f has no vertical asymptotes.
x-3
x-0
f has right horizontal asymptote y = 3/2 and left horizontal asymptote y = -3/2.
(b) Study the continuity and differentiability of the function, locate and classify possible points of
discontinuity and non-differentiability in dom f and compute its derivative.
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