   Chapter 3.1, Problem 55E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the points on the curve y = 2x3 + 3x2 – 12x + 1 where the tangent is horizontal.

To determine

To find: The points on the curve when the tangent is horizontal.

Explanation

Given:

The equation of the curve y=2x3+3x212x+1.

Derivative rules:

(1) Constant Function: ddx(c)=0

(2) Constant Multiple Rule: ddx[cf(x)]=cddxf(x)

(3) Power Rule: ddx(xn)=nxn1

(4) Sum Rule: ddx[f(x)+g(x)]=ddx(f(x))+ddx(g(x))

(5) Difference Rule: ddx[f(x)g(x)]=ddx(f(x))ddx(g(x))

Calculation:

Obtain the points on the curve if the tangents are horizontal.

The derivative of y is given below.

dydx=ddx(y) =ddx(2x3+3x212x+1)

Apply the sum rule (4) and difference rule (5),

dydx=ddx(2x3)+ddx(3x2)ddx(12x)+ddx(1)

Apply the derivative rules (2), (3) and (1),

dydx=2ddx(x3)+3ddx(x2)12ddx(x)+ddx(1)=2(3x31)+3(2x21)12(1x11)+0=6x2+6x12

Thus, the derivative of the curve y=2x3+3x212x+1 is 6x2+6x12

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