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 46E

(a)

To determine

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

Expert Solution

Answer to Problem 46E

The equation of the tangent line to the curve y=|x|2x2 at (1,1) is y=2x1_.

Explanation of Solution

Given:

The function is y=|x|2x2 and the point is (1,1).

Result used:

The Power Rule combined with the Chain Rule:

If n is any real number and g(x) is differentiable function, then

ddx[g(x)]n=n[g(x)]n1g(x) (1)

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 (2)

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (3)

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

Calculation:

For x>0,|x|=x and f(x)=x2x2.

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

dydx=ddx(y)=ddx(|x|2x2)=ddx(x2x2)

Apply the quotient rule as shown in equation (2),

dydx=2x2ddx[x]xddx[2x2][2x2]2

Apply the power rule combined with the chain rule as shown in equation (1),

dydx=2x2[1x11]xddx(2x2)12[(2x2)12]2=2x2[1]x12(2x2)121ddx(2x2)(2x2)=2x2x12(2x2)122(ddx(2)ddx(x2))(2x2)=2x2x12(2x2)12(02x21)(2x2)

On further simplification, the derivative of the function becomes,

 dydx=2x2+x2(2x2)12(2x)(2x2)=2x2+x22x2(2x2)=2x2+x22x2(2x2)=22x2(2x2)

Therefore, the derivative of y=|x|2x2 is dydx=22x2(2x2)_.

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

 m=dydx|x=1=22(1)2(2(1)2)    =21(1)  =2

Thus, the slope of the tangent line is m=2.

Substitute (1,1) for (x1,y1) and 2 for m in equation (1),

y1=2(x1)y1=2x2y=2x2+1y=2x1

Therefore, the equation of the tangent line is y=2x1.

(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=|x|2x2.

The equation of the tangent line is y=2x1.

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 46E

From Figure 1, it is observed that the equation of the tangent line touches on the curve y=|x|2x2 at the point (1,1).

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