An equation of the line tangent to the graph of f(x) = -x(1 – 2x)³ at the point (1,1) is a. y = -7x + 6 b. у %3D —6х + 5 с. у %3D —2х +1 d. y %3D 2х — 3 е. у%3D7х — 6 = -

Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
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Chapter1: Equations And Graphs
Section1.10: Modeling Variation
Problem 4E
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. An equation of the line tangent to the graph of f (x) = -x(1 –- 2x)³ at the point (1,1)
is
а. у %3D —7х+6
b. у 3D —6х + 5
С. у %3D —2х+ 1
d. y = 2x – 3
е. у%3D 7х — 6
Transcribed Image Text:. An equation of the line tangent to the graph of f (x) = -x(1 –- 2x)³ at the point (1,1) is а. у %3D —7х+6 b. у 3D —6х + 5 С. у %3D —2х+ 1 d. y = 2x – 3 е. у%3D 7х — 6
dy
If x2 + xy + y³ = 0, then, in terms of x and y,
dx
2х+y
a.
-
x+3y2
x+3y2
b.
2х+y
-2x
С.
1+3y2
-2x
d.
x+3y2
2х+y
е.
x+3y2-1
II
Transcribed Image Text:dy If x2 + xy + y³ = 0, then, in terms of x and y, dx 2х+y a. - x+3y2 x+3y2 b. 2х+y -2x С. 1+3y2 -2x d. x+3y2 2х+y е. x+3y2-1 II
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