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 5E
To determine

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

Expert Solution

Answer to Problem 5E

The equation of the tangent line to the curve y=4x3x2 at the point (2,4) is y=8x+12.

Explanation of Solution

Given:

The equation of the curve is y=4x3x2.

The curve passing through the point (2,4).

Formula used:

The slope of the tangent curve y=f(x) at the point P(a,f(a)) is,

m=limxaf(x)f(a)xa (1)

The equation of the tangent line to the curve y=f(x) at the point (a,f(a)) is,

yf(a)=f(a)(xa) (2)

Calculation:

Obtain the slope of the tangent line to the curve at the point (2,4).

Substitute a=2 and f(a)=4 in equation (1),

m=limx2f(x)f(2)x1=limx2(4x3x2)(4)x2=limx23x2+4x+4x2=limx2(3x24x4)x2

The factors of (3x24x4) is (3x+2) and (x2).

Thus, the slope of the tangent line to the curve becomes, m=limx2(3x+2)(x2)(x2).

Since the limit x approaches 2 but not equal to 2, cancel the common term (x2)(0) from both the numerator and the denominator,

m=limx2(3x2)=(3(2)2)=8

Thus, the slope of the tangent line to the curve at the point (2, −4) is, m=8_.

Obtain the equation of the tangent line.

Since the tangent line to the curve y=f(x) at (a,f(a)) is the line through the point (a,f(a)) whose slope is equal to the derivative of a, the value of f(a)=8.

Substitute a=2, f(a)=4 and f(a)=8 in equation (2),

(yf(a))=f(a)(xa)y(4)=8(x2)y+4=8x+16

Isolate y as shown below.

y=8x+164=8x+12

Thus, the equation of the tangent line is y=8x+12_.

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