   Chapter 2.3, Problem 77E

Chapter
Section
Textbook Problem
1 views

# Find the points on the curve y = 2 x 3 + 3 x 2 − 12 x + 1 where the tangent is horizontal.

To determine

To find:

The points on the curve y=2x3+3x2-12x+1 where the tangent is horizontal

Explanation

Concept:

Slope of tangent is equal to the derivative at the point. So horizontal tangent occurs when derivative is zero.

Formula:

y'=dydx

Constant function rule: d(C)dx=0

Power function rule: d(xn)dx=nxn-1

Constant multiple rule: d(Cf(x))dx=Cd(f(x))dx

Difference rule: d(fx-g(x))dx=d(f(x))dx-d(g(x))dx

Sum rule: d(fx+g(x))dx=d(f(x))dx+d(g(x))dx

Given:

y=2x3+3x2-12x+1

Calculation:

First find y'

y'=dydx=d(2x3+3x2-12x+1)dx

By sum and difference rule,

y'=d(2x3)dx+d(3x2)dx-d

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### f(x) = (x2 3)2 has points of inflection at x = _____. a) 0,3,3 b) 1, 1 c) 0, 1, 1 d) 3,3

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

#### Find for x = 3t2 + 1, y = t6 + 6t5. t4 + 5t3 4t3 + 15t2

Study Guide for Stewart's Multivariable Calculus, 8th

#### Solve each proportion. x-2x=x-65

College Algebra (MindTap Course List) 