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 3.4, Problem 45E

(a)

To determine

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

Expert Solution

Answer to Problem 45E

The equation of the tangent line to the curve y=2(1+ex) at (0,1) is y=x2+1.

Explanation of Solution

Given:

The function is y=2(1+ex).

Derivative Rule: Quotient Rule

If f(x). and g(x) are both differentiable function, then

ddx[f(x)g(x)]=g(x)ddx[f(x)]f(x)ddx[g(x)][g(x)]2 (1)

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (2)

where, m is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

Calculation:

The derivative of y is dydx, which is obtained as follows,

dydx=ddx(y)=ddx(2(1+ex))

Apply the quotient rule as shown in equation (1),

ddx[2(1+ex)]=(1+ex)ddx[2]2ddx[(1+ex)](1+ex)2=(1+ex)[0]2(ddx(1)+ddx(ex))(1+ex)2=02(0+(ex))(1+ex)2=2(ex)(1+ex)2

Therefore, the derivative of y=2(1+ex) is dydx=2(ex)(1+ex)2_.

The slope of the tangent line at (0,1) is computed as follows,

 m=dydx|x=02(e0)(1+e0)2    =2(1)(2)2       [Qe0=1]=12

Thus, the slope of the tangent line at (0,1) is m=12.

Substitute (0,1) for (x1,y1) and m=12 in equation (1),

(y1)=12(x0)y1=x2y=x2+1

Therefore, the equation of the tangent line to the curve y=2(1+ex) at (0,1) is y=x2+1.

(b)

To determine

To sketch: The graph of the curve and the tangent line.

Expert Solution

Explanation of Solution

Given:

The equation of the curve is y=2(1+ex).

The equation of the tangent line is y=x2+1.

Graph:

Use the online graphing calculator to draw the graph of the functions as shown below in Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.4, Problem 45E

From the Figure 1, it is observed that the equation of the tangent line touches on the curve y=2(1+ex) at the point (0,1).

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