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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directions at top of first page. please help with number 6.
Transcribed Image Text:PRACTICE:
For each function f(x), determine where f(x) is positive and where f(x) is negative.
Write your answers in interval notation.
1. f(x)=x²-15x +36
X²-12x-3x+36
=
= x(x-12) -3(x-12)
(x-12)(x-3)
f=0: x-12=0
X = 12
2. f(x) =
where is the fixn zero, where is the fxn undefined
f=0: x+4=0
(X=-4
x+4
x²(x-3)5
x-3=0
x = 3
f=undef: x²=0 (x-3) = 0
X=0°
X-3=0
X=3
-12-3
f= undef: never-polynomial
+
3
(-5) -4
positive: (-00, 3) u (12,00)
negative: (3, 12)
+
+
(2).
O
(۱)
TUR (-00,-4) v (3,00)
12
+
++
+
3 (4)
+
+
+
(13)
7
![5. f(x) = 2ex - 32e²x
f'(x) = 2x [e¹²] + x [32**]
= a(e** £[4x])+d [-32e²x]
= 2(e** (4.1)) + (-32e²g
= 8€ ²x + 1 + [32€²]
=&e 4²-32[e**] = 8e**- 32(e** & ax])
ax
ax
= 8e" ²-32(ex (2.1)) = 8e¹x²-3a(e²* a)
ax
f(x)=&e¹² - 64€²x = 0
f=0: 8e²³x (e²x-8)=0
870
bisog ei (2) merive enough
-liqueia si if
a ai if sooted espl
22x
2x=0 e³x8=0
in
In (e³x) - Ino €²x = 8
undefined in (@²x) = In (8)
No Sol'n
6. f(x) = (x+3)(x − 5)√√x – 3
2x In(e) = In (8)
2x · 1 = In(8)
2x = In (8)
x = In (8)
xylo ov
6
+x);
In(s) ot
2
F
positive : (In(8)
negative: (0o, Int
5+= (S+*)
*(1+2)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F27d35152-6917-4f16-a4e1-3cd886d8c64b%2F73da383a-b828-4cbd-b927-4203e7350e68%2F0nmg9v5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5. f(x) = 2ex - 32e²x
f'(x) = 2x [e¹²] + x [32**]
= a(e** £[4x])+d [-32e²x]
= 2(e** (4.1)) + (-32e²g
= 8€ ²x + 1 + [32€²]
=&e 4²-32[e**] = 8e**- 32(e** & ax])
ax
ax
= 8e" ²-32(ex (2.1)) = 8e¹x²-3a(e²* a)
ax
f(x)=&e¹² - 64€²x = 0
f=0: 8e²³x (e²x-8)=0
870
bisog ei (2) merive enough
-liqueia si if
a ai if sooted espl
22x
2x=0 e³x8=0
in
In (e³x) - Ino €²x = 8
undefined in (@²x) = In (8)
No Sol'n
6. f(x) = (x+3)(x − 5)√√x – 3
2x In(e) = In (8)
2x · 1 = In(8)
2x = In (8)
x = In (8)
xylo ov
6
+x);
In(s) ot
2
F
positive : (In(8)
negative: (0o, Int
5+= (S+*)
*(1+2)
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