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 6E
To determine

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

Expert Solution

Answer to Problem 6E

The equation of the tangent line to the curve y=x33x+1 at the point (2,3) is y=9x15.

Explanation of Solution

Given:

The equation of the curve is y=x33x+1.

The curve passing through the point (2,3).

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 parabola at the point (2,3).

Substitute a=2 and f(a)=3 in equation (1),

m=limx2f(x)f(2)x2=limx2(x33x+1)(3)x2=limx2x33x+13x2=limx2x33x2x2

The factors of (x33x2) is (x+1)2 and (x2).

Thus, the slope of the tangent line becomes, m=limx2(x+1)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(x+1)2=(2+1)2=32=9

Thus, the slope of the tangent line to the curve at the point (2, 3) is m=9_.

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)=9.

Substitute a=2,f(a)=3 and f(a)=9 in equation (2),

(yf(a))=f(a)(xa)y(3)=9(x2)y3=9x18

Isolate y as shown below.

y=9x18+3=9x15

Thus, the equation of the tangent line is y=9x15_.

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