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 8E
To determine

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

Expert Solution

Answer to Problem 8E

The equation of the tangent line to the curve y=2x+1x+2 at the point (1,1) is y=13x+23.

Explanation of Solution

Given:

The equation of the curve is y=2x+1x+2.

The curve passing through the point (1, 1).

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 (1, 1).

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

m=limx1f(x)f(1)x1=limx1(2x+1x+2)1x1=limx1(2x+1)1(x+2)(x+2)x1=limx1((2x+1)(x+2)(x1)(x+2))

=limx12x+1x2(x1)(x+2)=limx1x1(x1)(x+2)

Since the limit x approaches 1 but not equal to 1, cancel the common term (x1)(0) from both the numerator and the denominator,

m=limx11(x+2)=11+2=13

Thus, the slope of the tangent line to the curve at the point (1, 1) is m=13_.

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)=13.

Substitute a=1, f(a)=1 and f(a)=13 in equation (2),

(yf(a))=f(a)(xa)y(1)=13(x1)y1=13x13

Isolate y as shown below.

y=13x13+1=13x+23

Thus, the equation of the tangent line is y=13x+23.

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