Solve the problem. 23) Find the points where the graph of the function have horizontal tangents. f(x) = x3 - 6x A) (2, -12/2) C) (-V2, 4/2), (WZ, -WZ) B) (-2, 4), (~Z, -4) D) (-/2, 12/2), (0,0), (/z, -12/2) 24) Find an equation of the tangent to the curve f(x) = 2x2 - 2x +1 that has slope 2. A) y = 2x +1 B) у %3 2х - 1 С) у %3 2х +2 D) y = 2x
Solve the problem. 23) Find the points where the graph of the function have horizontal tangents. f(x) = x3 - 6x A) (2, -12/2) C) (-V2, 4/2), (WZ, -WZ) B) (-2, 4), (~Z, -4) D) (-/2, 12/2), (0,0), (/z, -12/2) 24) Find an equation of the tangent to the curve f(x) = 2x2 - 2x +1 that has slope 2. A) y = 2x +1 B) у %3 2х - 1 С) у %3 2х +2 D) y = 2x
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...
Related questions
Question
23-24

Transcribed Image Text:Solve the problem.
23) Find the points where the graph of the function have horizontal tangents.
f(x) = x3 - 6x
A) (2, -12/2)
C) (-V2, 4/2), (WZ, -WZ)
B) (-2, 4), (~Z, -4)
D) (-/2, 12/2), (0,0), (/z, -12/2)
24) Find an equation of the tangent to the curve f(x) = 2x2 - 2x +1 that has slope 2.
A) y = 2x +1
B) у %3 2х - 1
С) у %3 2х +2
D) y = 2x
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