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

### Single Variable Calculus: Concepts...

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

### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 3.1, Problem 50E
To determine

## To find:The equation of the tangent line to the curve at the given point

Expert Solution

The tangent is horizontal at x=1±136

### Explanation of Solution

Given:

The function is

y=x3+3x2+x+3

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=x3+3x2+x+3...................(1)

The derivative of a function

y=f(x)y=f(x)=dydx

Differentiating the equation (1) with respect to x

dydx=ddx(x3+3x2+x+3)dydx=3x2+6x+1

The tangent are horizontal

dydx=03x2+6x+1=0x=6±624(3)(1)2(3)x=6±36126x=6±246x=1±136

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