The arc length function for a curve y = f(x), where f is an increasing function, is s(x) = 7t+ 10 dt. (a) If f has y-intercept 6, find an equation for f. f(x) (7x+9) ارو (1/4) 24 + 7 (b) What point on the graph of f is 4 units along the curve from the y-intercept? State your answer rounded to 3 decimal places. (x, y) = ((1.081, 6.748) .)
The arc length function for a curve y = f(x), where f is an increasing function, is s(x) = 7t+ 10 dt. (a) If f has y-intercept 6, find an equation for f. f(x) (7x+9) ارو (1/4) 24 + 7 (b) What point on the graph of f is 4 units along the curve from the y-intercept? State your answer rounded to 3 decimal places. (x, y) = ((1.081, 6.748) .)
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section: Chapter Questions
Problem 18T
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Question
Hello, I am stuck on the second part of the problem and not sure where I am going wrong. I found f(x) then I set the antiderivative of s(x) to 4 and solve to get x. Once I get x, I plug it back into f(x) to find y but I guess I am missing something.
Transcribed Image Text:The arc length function for a curve y = f(x), where f is an increasing function, is s(x) =
(a) If f has y-intercept 6, find an equation for f.
f(x)=
2
(7x+9) (½³) 24
(7x+9
+
7
7t + 10 dt.
(b) What point on the graph of f is 4 units along the curve from the y-intercept? State your answer rounded to 3 decimal places.
(x, y)
= ((1.081, 6.748)
![7) y = f(x), withRE & IS A
[(4)
X
Wirende
1 + 5' (7)² = 7x+10
{f(x)]² = 7x+9
flick
fxd:mats
4-7779
Rele
flor=/ for cha? 1/3-622)
312
H
h=1415(4
sty-fu
системи
du=7dE
2
(0472)
12
T
3/
V
Soudnes = (3472)
62 [(7x+0) = (70940))
42
ム
21
(5+2) 끌어
2+2 (5+ + 2) =—2/25
2+2 (5+ (92) 12 = ?
7=1015
2
)+c
6= = (27) + c
21
6 = 54 +6
21
C-31-0)] 6-53-1
+1
72
12 = 24 =
2
1+2*-11-
2=
2
f(x) = 1 / (7x+9) + 274 |
21
5(x)=4
"
4 = 2²² (7x+16)² - 24 √10
4=류
21
+ Ju
4 + 20VTTU = 2 (7x+10) 31/2
+h
2
21
21 (4+247) = (7x+12) +2
2
42+ 16.JTU = (7x+10)
3
(42-1010)=7x+10
7x = (1/2 + 10570)
(42+1050)
X₁ =
(x, f(..)) = (1.081, 6.798)
〃
7
213
3-10
(3
13-10 ~11081](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b313955-89cd-4b2a-a000-54fbe962adf3%2F50b588ea-4906-44d0-8fc0-dd265b0d14fb%2Flatdtxg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:7) y = f(x), withRE & IS A
[(4)
X
Wirende
1 + 5' (7)² = 7x+10
{f(x)]² = 7x+9
flick
fxd:mats
4-7779
Rele
flor=/ for cha? 1/3-622)
312
H
h=1415(4
sty-fu
системи
du=7dE
2
(0472)
12
T
3/
V
Soudnes = (3472)
62 [(7x+0) = (70940))
42
ム
21
(5+2) 끌어
2+2 (5+ + 2) =—2/25
2+2 (5+ (92) 12 = ?
7=1015
2
)+c
6= = (27) + c
21
6 = 54 +6
21
C-31-0)] 6-53-1
+1
72
12 = 24 =
2
1+2*-11-
2=
2
f(x) = 1 / (7x+9) + 274 |
21
5(x)=4
"
4 = 2²² (7x+16)² - 24 √10
4=류
21
+ Ju
4 + 20VTTU = 2 (7x+10) 31/2
+h
2
21
21 (4+247) = (7x+12) +2
2
42+ 16.JTU = (7x+10)
3
(42-1010)=7x+10
7x = (1/2 + 10570)
(42+1050)
X₁ =
(x, f(..)) = (1.081, 6.798)
〃
7
213
3-10
(3
13-10 ~11081
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