# The equation of the tangent and parallel to the line

### 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 54E
To determine

## To find:The equation of the tangent and parallel to the line

Expert Solution

The equation of the both line is y=3x+12eln(3)

### Explanation of Solution

Given:

The tangent to the curve

y=1+2ex3x

Parallel to the line

3xy=5y=3x5

Concept used:

The equation is in slope −intercept form, y=mx+c

An equation for the line through the point (x1,y1) with slope m is

yy1=m(xx1)

Calculation:

The function

y=1+2ex3x.......................(1)

The derivative of a function

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

Differentiating the equation (1) with respect to x

y=1+2ex3xy=2ex3........................(2)

The derivative is slope of the tangent line so in order to the slope of the tangent line

The derivative of constant is zero

y=3x5y=3.............................(3)

The equation is in slope −intercept form

The slope of the line is =3

From equation (2) and equation (3)

2ex3=3ex=3x=ln(3)

The coordinates of points

y=1+2eln(3)3ln(3)

An equation for the line through the point (x1,y1) with slope m is

yy1=m(xx1)y1+2eln(3)3ln(3)=3(xln(3))y=3(xln(3))+12eln(3)+3ln(3)y=3x+12eln(3)

Draw the table

y=3x+12eln(3)

Test one point in each of the region formed by the graph

If the point satisfies the function then shade the entire region to denote that every point in the region satisfies the function

 x−axis 0 0.4 1.18 0.66 y−axis 3 2.78 4.02 4

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