BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.1, Problem 49E
To determine

To find:

The equation of the tangent line to the curve at the given point

Expert Solution

Answer to Problem 49E

The tangent is horizontal at (2,21) and (1,6)

Explanation of Solution

Given:

The function is

  y=2x3+3x212x+1

Concept used:

An equation of a line l through a point P1(x1,y1) with slope m . If P(x,y) is any point with xx1

Then P is on l if and only if the slope of the line through P1 and P

The equation is yy1=m(xx1)

Calculation:

The function

  y=2x3+3x212x+1...................(1)

The derivative of a function

  y=f(x)y=f(x)=dydx

Differentiating the equation (1) with respect to x

  dydx=ddx(2x3+3x212x+1)dydx=6x2+6x12

The tangent are horizontal

  dydx=06x2+6x12=0(x+2)(x1)=0x=2,1

Putting x=2,1 in equation (1)

  y(1)=2(1)3+3(1)212(1)+1y(1)=6y(2)=2(2)3+3(2)212(2)+1y(2)=21

The tangent is horizontal at (2,21) and (1,6)

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