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 2.6, Problem 26E

a.

To determine

To find:The equation of the tangent line to the curve G(x)=4x2x3 at the points (2,8)   and (3,9) .

Expert Solution

Answer to Problem 26E

The tangent line is y=3x+18 .

Explanation of Solution

Given: G(x)=4x2x3 at the point (2,8)   and (3,9) .

Concept used: G'(a)=limh0[G(x+h)G(x)]h

ExplanationLet G(x)=4x2x3 be any function.

It has to find the tangent line of the curve and derivative of the function G'(a)=limh0[G(x+h)G(x)]h

  G'(a)=limh0[4(x+h)2(x+h)3][4x2x3]hG'(a)=limh0[4(x2+h2+2xh)(x3+h3+3x2h+3xh2)][4x2x3]hG'(a)=limh0[4x2+4h2+8xhx3h33x2h3xh24x2+x]h=limh0[4h2+8xhh33xh(x+h)]h=limh04h+8xh23x(x+h)=8x3x2

Finding equation of tangent at

  (2,8)

Slope of tangent at

  (2,8)

  G'(2)=8(2)3(2)2=1612=4

Therefore the equation of the tangent is

  y(8)=4(x2)y8=(4x8)y4x=0y=4x

Finding equation of tangent at

  (3,9)

Slope of tangent at

  (3,9)

  G'(3)=8(3)3(3)2=2427=3

Therefore the equation of the tangent is

  y(9)=3(x3)y9=(3x+9)y+3x=18y=3x+18

Hence, the tangent line is y=3x+18 ..

b.

To determine

To illustrate: The graph of the curve and the tangent line on the same screen.

Expert Solution

Explanation of Solution

Given: G(x)=4x2x3 at the point (2,8)   and (3,9) .

Let y=5x/1+x2 be any function.

In the graph

The black curve is y=4x2x3

The blue line which is tangent at (3,9) has equation y=3x+18

The red line which is tangent at (2,8) has equation y=4x

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 2.6, Problem 26E

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